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Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 7th Lesson Moving Charges and Magnetism Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 7th Lesson Moving Charges and Magnetism

Very Short Answer Questions

Question 1.
What is the importance of Oersted’s experiment ?
Answer:
Importance of Oersted’s experiment is every current carrying conductor produces a magnetic field around it and which is perpendicular to current carrying conductor.

Question 2.
State Ampere’s law and Biot-Savart law. (T.S. Mar. ’19)
Answer:
Ampere’s law : The line integral of the intensity of magnetic induction around a closed path is equal to µ₀ times the total current enclosed in it.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = \(\mu_0 \mathrm{i}\)

Biot – Savart’s laws : Biot – Savart’s states that the intensity of magnetic induction (dB) due to a small element is directly proportional to the .

1. current (i)
2. length of the element (dl)
3. sine angle between radius vector (r) and dl and
4. Inversely proportional to the square of the point from current element.
   ∴ dB ∝ \(\frac{\mathrm{i} \mathrm{d} l \sin \theta}{\mathrm{r}^2}\)
   dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i} d l \sin \theta}{r^2}\)

Question 3.
Write the expression for the magnetic induction at any point on the axis of a circular current-carrying coil. Hence, obtain an expression for the magnetic induction at the centre of the circular coil.
Answer:

1. Intensity of magnetic induction field on the axis of the circular coil B = \(\frac{\mu_0 n i r^2}{2\left(r^2+x^2\right)^{3 / 2}}\)
2. At the centre of the coil B = \(\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}\)

Question 4.
A circular coil of radius V having N turns carries a current “i”. What is its magnetic moment ?
Answer:
Magnetic moment (M) = N i A
M = N i (πr²) (∵ A = πr²)
∴ M = n N i r²

Question 5.
What is the force on a conductor of length L carrying a current “i” placed in a magnetic field of induction B ? When does it become maximum ?
Answer:

1. Force on a conductor (F) = B i L sin θ
2. If θ = 90°, F_Max = B i L
   i.e., the direction of current and magnetic field are perpendicular to each other, then force is maximum.

Question 6.
What is the force on a charged particle of charge “q” moving with a velocity “v” in a uniform magnetic field of induction B ? When does it become maximum ?
Answer:

1. Force on a charged particle (F) = B q v sin θ.
2. If θ = 90°, F_Max = B q v.

Question 7.
Distinguish between ammeter and voltmeter. (A.P. Mar. ’15)
Answer:
Ammeter

1. It is used to measure current.
2. Resistance of an ideal Ammeter is zero.
3. It is connected in series in the circuits.

Voltmeter

1. It is used to measure P.D between two points.
2. Resistance of ideal voltmeter is infinity.
3. It is connected in parallel in the circuits.

Question 8.
What is the principle of a moving coil galvanometer ?
Answer:
Moving coil galvanometer is based on the fact that when a current carrying coil is placed in a uniform magnetic field, it experiences a torque.
∴ current in the coil (i) ∝ deflecting angle (θ).

Question 9.
What is the smallest value of current that can be measured with a moving coil galvanometer ?
Answer:
Moving coil galvanometer is sensitive galvanometer, it is used to measure very small current upto 10^-9A.

Question 10.
How do you convert a moving coil galvanometer into an ammeter ? (A.P. Mar. ’19)
Answer:
A small resistance is connected in parallel to the moving coil galvanometer, then it converts to Ammeter.
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)

Question 11.
How do you convert a moving coil galvanometer into a voltmeter? (T.S. & A.P. Mar. ’16, T.S. Mar. ’15 Mar. ’14)
Answer:
A high resistance is connected in series to the moving coil galvanometer, then it converts to voltmeter.
R = \(\frac{\mathrm{v}}{\mathrm{i}_{\mathrm{g}}}-\mathrm{G}\)

Question 12.
What is the relation between the permittivity of free space e₀, the permeability of free space m₀ and the speed of light in vacuum ?
Answer:
Speed of light in vacuum (C) = \(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Here μ₀ = m₀ = permeability in vacuum
ε₀ = permittivity in vacuum.

Question 13.
A current carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns about the vertical axis ?
Answer:
Torque \((\tau)\) = \(\overrightarrow{\mathrm{M}}\) × \(\overrightarrow{\mathrm{B}}\) = i\(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) (M = n i A)
where i is current, \(\overrightarrow{\mathrm{A}}\) is area vector, \(\overrightarrow{\mathrm{B}}\) is magnetic field. Area vector \(\overrightarrow{\mathrm{A}}\) acts normal to the loop, so torque \(\vec{\tau}\) cannot act along the vertical axis. The magnetic field is not set up to turn the loop around it self.

Question 14.
A current carrying circular loop is placed in a uniform external magnetic field. If the loop is free to turn, what is its orientation when it is achieves stable equilibrium ?
Answer:
The plane of the loop is perpendicular to the direction of magnetic field because the torque on the loop in this orientation is zero.

Question 15.
A wire loop of Irregular shape carrying current is placed In an external magnetic field. If the wire is flexible, what shape will the loop change to ? Why?
Answer:
For a given perimeter, a circle has maximum area among all geometrical shapes. So to maximise tie magnetic flux through it will assume a circular shape with its plane normal to the field.

Short Answer Questions

Question 1.
State and explain Biot-Savart law. (T.S. Mar. ’16, Mar. ’14)
Answer:
Consider a very small element of length dl of a conductor carrying current

Magnetic induction due to small element at a point P distance r form the element. Magnetic induction (dB) is directly proportional to

1. 1. current (i)
   2. Length of the element (dl)
   3.  sine angle between r and dl and
   4. Inversely proportional to the square of the distance from small element to point R
      
      dB ∝ \(\frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
      dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
      where μ = permeability in free space.
      \(\frac{\mu_0}{4 \pi}\) = 10^-7 Wb m^-1 A^-1

Question 2.
State and explain Ampere’s law.
Answer:
Ampere’s law : The line integral of the intensity of magnetic induction field around closed path is equal to μ₀ times the net current (i) enclosed by the path.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i
Proof: Consider a long” straight conductor carrying current i as shown in figure. Magnetic induction at a distance r from the conductor is given by

B = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) (from Biot-Savart’s law)
The value of B is same at all points on the circle.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = \(\oint \mathrm{Bd} l \cos \theta\)
= \(\mathrm{B} \oint \mathrm{d} l\) = B × 2πr (∵ θ = 0° Angle between \(\overrightarrow{\mathrm{B}}\) & \(\overrightarrow{\mathrm{d} l}\) is zero)
= \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\) × 2πr
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = μ₀i.
This proves Ampere’s circuital laws.

Question 3.
Find the magnetic induction due to a long current carrying conductor.
Answer:
Consider a long straight conductor carrying a current i. Let P be a point at a distance r from the conductor. Let r be the radius of the circle passing through point p.

(∵ \(\oint \mathrm{d} l\) = 2πr = circumference of the circle)
Magnetic induction is same at all points on the circle. Consider a small element of length dl.
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = \(\oint \mathrm{Bd} l \cos \theta\)
Angle between B and dl is zero i.e. θ = 0
= B\(\oint \mathrm{dl}\)
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}\) = B (2πr) ——> (1)
According to Ampere’s laws
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = µ₀i ——-> (2)
From equations (1) and (2), B (2πr) = µ₀ i
B = \(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}\)

Question 4.
Derive an expression for the magnetic induction at the centre of a current carrying circular coil using Biot-Savart law.
Answer:
Consider a circular coil of radius r and carry a current i. Consider a small element ‘dl’. Let O is the centre of the coil. By using Biot – Savart’s law
dB = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i} d l \sin \theta}{\mathrm{r}^2}\)
Here angle \(\overrightarrow{\mathrm{d} t}\) and \(\overrightarrow{\mathrm{r}}\) is 90° (i.e. θ = 90°)
dB = \(\frac{\mu_0}{4 \pi} \frac{\mathrm{id} l}{\mathrm{r}^2}\) — (1)

As the field due to all elements of the circular loop have the same direction.
The resultant magnetic field can be obtained by integrating equation (1)
\(\int \mathrm{dB}\) = \(\int \frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l}{\mathrm{r}^2}\)
B = \(\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}^2} \int \mathrm{d} l\) ∵ (\(\int\)dl = 2πr)
= \(\frac{\mu 0 \mathrm{i}}{4 \pi \mathrm{r}^2}\) × 2πr
B = \(\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}}\)
If the circular coil has n turns.
B = \(\frac{\mu_0 \mathrm{ni}}{2 \mathrm{r}}\)

Question 5.
Derive an expression for the magnetic induction a point on the axis of a current carrying circular coil using
Answer:

Consider a circular coil of radius R and carrying a current i. Let P is a point on the axis at a distance x from the centre O. Let r be the distance of small element (dl) from P.
From Biot – savart’s law
dB = \(\frac{\mu_0}{4 \pi} \frac{\text { id } l \sin \theta}{r^2}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l}{\mathrm{r}^2}\) —– (1) (∵ θ = 90° Angle between \(\overrightarrow{\mathrm{d} l}\) and \(\overrightarrow{\mathrm{r}}\))
dB can be resolved into two components dB cos θ and dB sinθ. If we consider another element diametrically opposite to AB.
This also resolved into dB cosθ and dB sinθ.
The components along the axis will add up and perpendicular to the axis will cancel.
∴ Resultant magnetic induction at P is
B = \(\int \frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{id} l \sin \theta}{\mathrm{r}^2}\)
= \(\frac{\mu_0 i}{4 \pi^2} \int d / \sin \theta\) (∵ sin θ = \(\frac{R}{r}\))
= \(\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}^2}\) × 2πR × \(\frac{\mathrm{R}}{\mathrm{r}}\) (∵ \(\int dl\) = 2πr)
B = \(\frac{\mu_0 \mathrm{iR}^2}{2 \mathrm{r}^3}\) —–> (3)
From figure r = \(\sqrt{\mathrm{R}^2+\mathrm{x}^2}\)
B = \(\frac{\mu_0 i R^2}{2\left(R^2+x^2\right)^{3 / 2}}\) —–> (4)
If the coil contains N turns,
B = \(\frac{\mu_0 N i R^2}{2\left(R^2+x^2\right)^{3 / 2}}\) —–> (5)

Question 6.
Obtain an expression for the magnetic dipole moment of current loop.
Answer:
We know that magnetic induction on the axial line of a circular coil is B = \(\frac{\mu_0 N \mathrm{iR}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{3 / 2}}\)
where N = Number of turns in the coil
R = Radius of the coil
x = Distance from centre of the coil
i = Current in a coil
If x >> R, Then B = \(\frac{\mu_0 \mathrm{Ni} \mathrm{R}^2}{2 \mathrm{x}^3}\)
Multiplying and dividing with 2π
B = \(\frac{\mu_0 \mathrm{Ni} \mathrm{R}^2}{2 \mathrm{x}^3} \times \frac{2 \pi}{2 \pi}\)
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{Ni}\left(\pi \mathrm{R}^2\right)}{\mathrm{x}^3}\) (∵ A = πR²)
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{Ni} \mathrm{A}}{\mathrm{x}^3}\) —– (1)
We know that magnetic induction field on the axial line of a bar magnet
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{M}}{\mathrm{x}^3}\) —– (2)
Comparing the equations (1) and (2)
Magnetic moment (M) = N i A

Question 7.
Derive an expression for the magnetic dipole moment of a revolving electron. (Ã.P. Mar. ’16)
Answer:
Consider an electron revolving in a circular orbit of radius r with speed v and frequency υ.
If the electron cross a point P on the circle in every revolution, then distance travelled by electron to complete one revolution = 2πr.
No. of revolutions in one second (υ) = \(\frac{\mathrm{v}}{2 \pi r}\)
The electric current (i) = \(\frac{\text { charge }}{\text { time }}\) = charge × frequency
i = e × \(\frac{\mathrm{v}}{2 \pi \mathrm{r}}\)
∴ Magnetic dipole moment (M) = I A (∵ N = 1)
M = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^2\) (∵ A = πr²)
M = \(\frac{\text { evr }}{2}\)

Question 8.
Explain how crossed E and B fields serve as a velocity selector.
Answer:
When a charged particle q moving with a velocity v in presence of both electric and magnetic fields.
The force experienced due to electric field F_E = q\(\overrightarrow{\mathrm{E}}\)
The force experienced due to magnetic field F_B = q \(\left(\begin{array}{l}
\vec{v} \times \vec{B}
\end{array}\right)\)
Consider electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle.
E = E\(\hat{j}\), B = B\(\hat{k}\), v = v\(\hat{i}\)
F_E = qE\(\hat{j}\), F_B = q (v\(\hat{i}\) × B\(\hat{k}\)) = – qvB\(\hat{j}\)
∴ F = F_E + F_B
F = q (E – υB) \(\hat{j}\)

Thus electric and magnetic forces are in opposite directions.
We adjust E and B such that, the forces are equal
F_E = F_B
qE’ = q υ B
υ = \(\frac{E}{B}\)
This condition can be used to select charged particles of a particular velocity. The crossed field E and B serve as a velocity selector.

Question 9.
What are the basic components of a cyclotron ? Mention its uses ?
Answer:
Cyclotron is a device used to accelerate positively charged particles like protons, α – particles, deutrons etc.

Cyclotron mainly consists of

1. Two hollow D-shaped metallic chambers D₁ and D₂
2. High frequency oscillator
3. Strong electro magnet
4. Vacuum chamber.

Uses of cyclotron :

1. It is used for producing radioactive material for medical purposes i.e. diagnostics and treatment of chronic diseases.
2. It is used to improve the quality of solids by adding ions.
3. It is used to synthesise fresh substances.
4. It is used to bombard the atoms with highly accelerated particles to study the nuclear reactions.

Long Answer Questions

Question 1.
Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Derive an expression for the force per unit length between two parallel current carrying conductors.
Answer:
Expression for the Force acting on a current carrying conductor :
Consider a straight conductor (wire) of length ‘l’, area of cross section ‘A’, carrying a current ’i’, ‘which is placed in a uniform magnetic field of induction B’ as shown in fig.
We know the external magnetic field exerts a force on the conductor.
The electrons in effect move with an average velocity called drift velocity V_d‘. The direction of conventional current will be opposite to the direction of drift velocity.

Let us assume that the current flows through the conductor from left ‘B’ in the plane of the paper makes an angle ‘θ’ with the direction of current ‘i’ as shown in fig.

If F’ is the force acting on the charge ‘q’ in B
∴ F = q V_d B sin θ
If ‘n’ represents number of moving electrons per unit volume (∵ n = \(\frac{\mathrm{N}}{\mathrm{V}}\))
∴ Current i = nqV_dA
If ‘N’ is the number of electrons in the length ‘l’
N = nlA .
Total force on conductor F = F’.N (∵ N = nV = n × A × l)
= (q V_dB sin θ) (nlA)
= (nqV_dA) (IB sin θ)
∴ F = ilB sin θ
Case (i) : If θ = 0°, F_Min = 0
Case (ii) : If θ = 90°, F_Max = Bil

Expression for the force between two Parallel conductors carrying conductors:
Consider two straight parallel conductors .‘AB and ‘CD’ carrying currents ‘i₁’ and “i₂’ and which are separated by a distance ‘r’ as shown in fig.

If B₁ and B₂ are magnetic inductions produced by the current carrying conductors AB and CD. Magnetic induction B₁ at a distance ‘r’ from the conductor ‘AB’ can be written as B₁ = \(\frac{\mu_0 i_1}{2 \pi r}\)
If ‘F is force acting on CD’ due to magnetic induction B₁ then
F_CD = i₂lB₁
Where l = length of the conductor
F_CD = \(\mathrm{i}_2 l\left(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\right)\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\) — (1)

The direction of the force can be determined by using Flemings left hand rule.
Similarly we can find the force acting on the conductor AB due to magnetic induction B₂.
F_AB = i₁lB₂
∴ F_AB = \(\mathrm{i}_1 l\left(\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{r}}\right)\) —– (2) [∵ B₂ = \(\frac{\mu_0 i_2}{2 \pi r}\)]
From the equations (1) and (2) F_AB = F_CD = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
∴ Force between two parallel, straight conductors carrying currents
F = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
Force per unit length \(\frac{\mathrm{F}}{l}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)

Question 2.
Obtain an expression for the torque on a current carrying loop placed in a uniform magnetic field. Describe the construction and working of a moving coil galvanometer.
Answer:
Torque acting on a coil carrying a current kept in a uniform magnetic field : Let a rectangular current loop ABCD of length l = AB = CD and width b = AD = BC carrying a current “i” be suspended in a magnetic field of flux density B.
The normal ON drawn to the plane of the coil makes an angle ‘θ’ with the magnetic field B.

Force on arm AD = i\(\overline{\mathrm{b}}\) × \(\overrightarrow{\mathrm{B}}\) acting upwards along the axis of suspension
Force on arm BC = i\(\overline{\mathrm{b}}\) × \(\overrightarrow{\mathrm{B}}\) acting downwards along the axis of suspension
Hence these two forces cancel.

Force on arm AB = ilB acting perpendicular to the plane as shown.
Force on arm CD = ilB acting perpendicular to the plane as shown.
These two forces constitute a couple on the coil.
Moment of the couple = (Force) × (Perpendicular distance between the forces) = i/B (PQ sin θ)
Torque = ilB b sinθ
But l × b = Area of coil
∴ Torque = iAB sin θ
If the loop has n’ turns the torque on the coil
\(\tau\) = n i AB sinθ
If “ϕ’ is the deflection of the coil, that is the angle between the plane of the coil and magnetic field B
\(\tau\) = n i AB cos ϕ

Moving coil galvanometer:
Principle : When a current carrying coil is placed in the uniform magnetic field, it experiences a torque.

Construction: .

1. It consists of a coil wound on a non metallic frame.
2. A rectangular coil is suspended between two concave shaped magnetic poles with the help of phosphour Bronze wire.
3. The lower portion of the coil ‘is connected to a spring.
4. A small plane mirror M is fixed to the phosphour Bronze wire to measure the deflection of the coil.
5. A small soft iron cylinder is placed with in the coil without touching the coil. ‘The soft iron cylinder increases the induction field strength.
6. The concave shaped magnetic poles render the field radial. So maximum torque acting on it.
7. The whole of the apparatus is kept inside a brass case provided with a glass window.
   

Theory:
Consider a rectangular coil of length l and breadth b and carrying current i suspended in the induction field strength B.
Deflecting torque \((\tau)\) = B i A N —–> (5)

where A = Area of the coil
N = Total number of turns.
The restoring torque developed in the suspension = C θ —-> (2)
Where C is the couple per unit twist and θ is the deflection made by the coil. When the coil is in equilibrium position
Deflecting torque = Restoring torque
B i A N = C θ
i = \(\left(\frac{C}{\mathrm{BAN}}\right) \theta\)
Where K = \(\frac{\mathrm{C}}{\mathrm{BAN}}\) = Galvanometer constant
i = K θ —-> (3)
i ∝ θ
Thus deflection of the coil is directly proportional to the current flowing through it.
The deflection in the coil is measured using lamp and scale arrangement.

Question 3.
How can a galvanometer be converted fo an ammeter ? Why is the parallel resistance smaller that the galvanometer resistance ?
Answer:
Conversion of Galvanometer into Ammeter :
Galvanometer is converted into an ammeter by connecting a suitable resistance is parallel to it.
This arrangement decreases the effective resistance.

Ammeter is used for measuring the current in an electric circuit and it is connected in series in circuit.
The inclusion of the ammeter in the circuit should not alter the current or total resistance of the circuit so it has very low resistance.
The resistance of An ideal Ammeter is zero.
Let G and S be the Galvanometer and shunt resistances respectively.
Let ‘i’ be the total current, divided at A into i_g and i_s as shown in fig.
From Kirchhoff’s I^st law i = i_g + i_s
As ‘G’ and ‘S’ are parallel
PD. across Galvanometer = P.D. across shunt
i_g G = i_s S
S = \(\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{i}_{\mathrm{s}}} \mathrm{G}\)
= \(\frac{\mathrm{Gi}_{\mathrm{g}}}{\mathrm{i}-\mathrm{i}_{\mathrm{g}}}\) [∵ i_s = i – i_g]
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)
If \(\frac{\mathrm{i}}{\mathrm{i}_g}\) = n ⇒ i_g = \(\frac{i}{n}\)
∴ The current flowing through the galvanometer be \(\left(\frac{1}{n}\right)^{\text {th }}\) of total current.
∴ S = \(\frac{\mathrm{G}}{\mathrm{n}-1}\)
If ‘R is the effective resistance between points ‘A’ and ‘B’ then

Hence current through galvanometer is proportional to the total current. Since ‘S’ is small major portion of the current flows through it and a small portion of current flows through G. So shunt protects the galvanometer from high currents. Parallel resistance is smaller than Galvanometer resistance because to protect the Galvanometer from high (large) current (or) to pass. Large currents through shunt and small current passes through the galvanometer.

Question 4.
How can a galvanometer be converted to a voltmeter ? Why is the series resistance greater that the galvanometer resistance ?
Answer:
Conversion of Galvanometer into, Voltmeter: A galvanometer-is converted into voltmeter by connecting a high resistance (R) in series with it. Voltmeter is used to measure the RD. between any two points in circuit and it is connected in parallel to the component of the circuit.

Let V be the potential difference to be measured between the points ‘A’ and ‘B’.
∴ V = (R + G) i_g [∴ V = iR]

Where G = Galvanometer Resistance
i_g = Current passing through the galvanometer
\(\frac{\mathrm{V}}{\mathrm{i}_g}\) = R + G
∴ R = \(\frac{\mathrm{V}}{\mathrm{i}_g}\) – G —- (1)
The value of ‘R’ can be calculated by using the above formula. If V_g is the maximum P.D. across the galvanometer then V_g = i_g G
∴ i_g = \(\frac{V_g}{G}\) —— (2)
Substitute ‘i_g‘ in Equ (1)
R = \(\frac{V G}{V_g}\) – G = G\(\left(\frac{V}{V_g}-1\right)\)
If \(\frac{V}{V_g}\) = n ⇒ R = G(n – 1)
Note: n = \(\frac{V}{V_g}\) is the ratio of maximum voltage to be measured to the maximum voltage across the galvanometer
Series resistance is greater than galvanometer resistance because the current in external resistance and potential difference will be decreased and to increase the resistance of the galvanometer.

Question 5.
Derive an expression for the force acting between two very long parallel current-carrying conductors and hence define the Ampere.
Answer:
Force between two parallel conductors carrying current:

Let two long parallel conductors A and B seperated by a distance r carry currents i₁ and i₂ in the same directions.
The current i₁ produces magnetic induction around the conductor A and the current i₂ produces magnetic induction B₂ around the conductor B.
If l is the length of each conductor,
The magnetic induction B₁ at distance r from conductor A is
B₁ = \(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\) — (1)
The force on conductor B is given by
F₂ = i₂ l B₁ —- (2)
The direction of force is given by Fleming left hand rule
F₂ = i₂l × \(\frac{\mu_0 \mathrm{i}_1}{2 \pi \mathrm{r}}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\) —- (3)
The direction of force F₂ is towards conductor A. Similary the magnetic induction B₂ at distance r from conductor B is
B₂ = \(\frac{\mu_0 \mathrm{i}_2}{2 \pi \mathrm{r}}\) — (4)
The force on conductor A is given by
F₁ = i₁lB₂ —- (5)
Substituting B₂ value in equation (5)
F₁ = \(\frac{\mathrm{i}_1 l \times \mu_0 \mathrm{i}_2}{2 \pi \mathrm{r}}\)
F₁ = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\) — (6)
It can be seen that | F₁| = | F₂| = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 l}{2 \pi \mathrm{r}}\)
The force per unit length of the conductor is given by
\(\frac{F}{l}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\) —–>(7)

Definition of Ampere:
If i₁ = i₂ = 1A, r = 1m
F/l = \(\frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1}\) = 2 × 10^-7 Nm^-1
“When two infinitely long parallel conductors, carrying the same current are seperated by 2 × 10^-7 Nm^-1, then the current flowing through each conductor is said to be one Ampere”.

Problems

Question 1.
A current of 10A passes through two very long wires held parallel to each other and separated by a distance of 1m. What is the force per unit length between them ? (T.S. Mar. ’19 & A.P. & T.S. Mar. ’15)
Solution:
i₁ = i₂ = 10A
r = 1m
\(\frac{\mathrm{F}}{l}\) = \(\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{r}}\)
= \(\frac{4 \pi \times 10^{-7} \times 10 \times 10}{2 \pi \times 1}\)
\(\frac{\mathrm{F}}{\mathrm{l}}\) = 2 × 10^-5 Nm^-1.

Question 2.
A moving coil galvanometer can measure a current of 10^-6 A. What is the resistance of the shunt required if it is to measure 1A ?
Solution:
Galvanometer current (i_g) = 10^-6A
i = 1A
S = \(\frac{\mathrm{G}}{\frac{\mathrm{i}}{\mathrm{i}_{\mathrm{g}}}-1}\)
= \(\frac{\mathrm{G}}{\frac{1}{10^{-6}}-1}\)
S = \(\frac{G}{10^6-1}\)
S = \(\frac{\mathrm{G}}{99,999} \Omega\)
Where G = Galvanometer resistance.

Question 3.
A circular wire loop of radius 30 cm carries a current of 3.5 A. Find the magnetic field at a point on its axis 40 cm away from the centre.
Solution:
Radius (r) = 30 cm = 30 × 10^-2m
Current (i) = 3.5 A
x = 40 cm = 40 × 10^-2m

Textual Exercises

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil ?
Solution:
Here, n = 100, r = 8cm = 8 × 10^-2 m and I = 0.40 A
The magnetic field B at the centre
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi \mathrm{In}}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 3.14 \times 0.4 \times 100}{8 \times 10^{-2}}\) = 3.1 × 10^-4T
The direction of magnetic field depends on the direction of current if the direction of current is anticlockwise. According to Maxwell’s right hand rule, the direction of magnetic field at the centre of coil will be perpendicular outwards to the plane of paper.

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire ?
Solution:
Here, I = 35 A and r = 20 cm = 0.2 m
The wire is along and it is considered as an infinite length wire. The magnetic field
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 35}{0.2}\) = 3.5 × 10^-5T

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Solution:
Given I = 50A and r = 2.5 m
The magnitude of magnetic field

B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) = 10^-7 × \(\frac{2 \times 50}{2.5}\)
= 4 × 10^-6T
The direction of magnetic field at point P is given by Maxwell’s right hand rule.

Question 4.
A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line ?
Solution:

Given I = 90 A and r = 1.5m
The magnitude of magnetic field
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 90}{1.5}\)
= 1.2 × 10^-5 T
The direction of magnetic flux is given by Maxwell’s right hand rule. So, the direction of magnetic field at point P due to the following current is perpendicularly outwards to the plane of paper.

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 300 with the direction of a uniform magnetic field of 0.15.T?
Solution:
According to the question
I = 8 A, θ = 30°, B = 0.15 T, l = 1 m
The magnitude of magnetic force

f = I(l × B) = I l B sin θ
= 8 × 1 × 0.15 × sin 30°
= \(\frac{8 \times 0.15}{2}\) = 4 × 0.15 = 0.6 N/m

Question 6.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire ?
Solution:
Here, the angle between the magnetic field and the direction of flow of current is 90°. Because the magneitc field due to a solenoid is along the axis of solenoid and the wire is placed perpendicular to the axis.

Given, l = 3 cm = 3 × 10^-2 m, I = 10A, B = 0.27 T
f = I l B sin 90°
= 10 × 3 × 10^-2 × 0.27 = 8.1 × 10^-2 N
According to right hand palm rule, the direction of magnetic force is perpendicular to plane of paper inwards.

Question 7.
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Given I₁ = 8A, I₂ = 5A and r = 4 cm = 0.4m
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}}\) = \(\frac{10^{-7} \times 2 \times 8 \times 5}{0.04}\) = 2 × 10^-4N
The force on A of length 10 cm is F¹ = F × 0.1 (∵ 1m = 100 cm)
F¹ = 2 × 10^-4 × 0.1
F¹ = 2 × 10^-5N.

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. It the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Solution:
The length of solenoid, l = 80 cm = 0.8 m
Number of layers = 5
Number of turns per layer = 400

Diameter of solenoid = 1.8 cm
Current in solenoid = I = 8 A
∴ The total number of turns N = 400 × 5 = 2000
and no. of turns / length, n = \(\frac{2000}{0.8}\) = 2500
The magnitude of magnetic field inside the solenoid
B = μ₀nl = 4 × 3 .14 10^-7 × 2500 × 8
= 2. 5 × 10^-2 T
The direction of magnetic field is along the axes of solenoid.

Question 9.
A square coil of side 10 cm consists of 20 turns and carries current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil ?
Solution:
Given, side of square coil = 10 cm = 0.1 m
Number of turns (n) = 20
current in square coil I = 12 A
Angle made by coil θ = 30°
Magnetic filed B = 0.80 T

The magnitude of torque experienced by the coil
\(\tau\) = NIAB sinθ
= 20 × 12 × (10 × 10^-2)² × 0.80 × sin 30°
= 0.96 N – m

Question 10.
Two moving coil meters, M₁ and M₂ have the following particulars :
R₁ = 10 Ω, n₁ = 30,
A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T
R₂ = 14Ω, n₂ = 42,
A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T, K₁ = K₂
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Solution:
Given, R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T
R₂ = 140Ω, n₂ = 42, A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T, K₁ = K₂

Question 11.
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
(e = 1.5 × 10^-19 C, m_e = 9.1 × 10^-31 kg)
Solution:
Given, magnetic field B = 6.5 G = 6.5 × 10^-4 T
Charge e = -1.6 × 10^-19 C
Speed of electron V = 4.8 × 10⁶ m/s
Mass of electron m_e = 9.1 × 10^-31 kg
Angle between magnetic field and electron (θ) = 90°
The force on charge particle entering in the magnetic field
F = q (V × B) = e (V × b)
The electron attains a circular path and necessarily centripetal force is provided by magnetic force.
e (V × B) = \(\frac{m V^2}{r}\)
e V B sin 90° = \(\frac{\mathrm{m} \mathrm{V}^2}{\mathrm{r}}\)
r = \(\frac{\mathrm{mV}}{\mathrm{e} \mathrm{B} \times 1}\) = \(\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}\) = 4.2 × 10^-2m = 4.2 cm

Question 12.
From Exercise 11 data, obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron ? Explain.
Solution:
Given
B = 6.5 G = 6.5 × 10^-4T, V = 4.8 × 10⁶ m/s, e = 1.6 × 10^-19C
m_e = 9.1 × 10^-3 kg
\(\frac{\mathrm{mV}^2}{\mathrm{r}}\) = q V B ⇒ \(\frac{\mathrm{mV}}{\mathrm{r}}\) = qB
If angular velocity of electron is ω, then
V = r ω
ω = \(\frac{\mathrm{qB}}{\mathrm{m}}\)
2πn = \(\frac{\mathrm{qB}}{\mathrm{m}}\) ⇒ n = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
Frequency of revolution of electron in orbit
υ = \(\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}\) = \(\frac{\mathrm{Be}}{2 \pi \mathrm{m}_{\mathrm{e}}}\) = \(\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}\) = 18.18 × 10⁶ Hz.

Question 13.
(a) Circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniiorm horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil form turning.
(b) Would your answer change, If the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area ? (All other particular are also unaltered.)
Answer:
a) Given, no. of turns (n) = 30, radius (r) = 8 cm = 0.08 m
Current in coil (I) = 6A, Magnetic field (B) = 1.0 T,
Angle made by field with the normal of the coil, θ = 60°
\(\tau\) = n I A Bsinθ
= 30 × 6 × π (0.08)² × 1 × sin 60°
= 30 × 6 × 3.14 × 0.08 × 0.08 × \(\frac{\sqrt{3}}{2}\)
\(\tau\) = 3.133 N – m

b) From the formula, it is clear that the torque on the loop does not depend on the shape if area remains constant. So, the torque remains constant.

Additional Exercises

Question 1.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X Is anticlockwise, and clockwise In Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Solution:
For coil X
Radius of coil, r_x = 16 cm = 0.16 m
No. of turns n_x = 20
Current in the coil I_x = 16A (Anti clockwise) .
For coil Y
Radius of coil, r_y = 10 cm = 0.1 m
No. of turns n_y = 25
Current in the coil I_y = 18 A (clockwise)
The magnitude of the magnetic field at the centre of coil X
B_x = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_x \cdot \pi n_x}{r_x}\) = \(\frac{10^{-7} \times 2 \times 16 \times 3.14 \times 20}{0.16} \mathrm{~T}\)
B_x = 4π × 10^-4T
The magnitude field at the centre B =B_y – B_x
= (9π – 4π) × 10^-4 = 5π × 10^-4
= 1.6 × 10^-3T (towards west)

Question 2.
A magnetic field of 1oo G (1 G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^-3 m². The maximum current carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a côre is at most 1000 turns m^-1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Solution:
Magnetic field B = 100 G = 100 × 10^-4T = 10^-2 T
Maximum current I = 15A, n = 1000/m.
The magnitude of magnetic field B = μ₀ nI
nI = \(\frac{\mathrm{B}}{\mu_0}\) = \(\frac{10^{-2}}{4 \times 3.14 \times 10^{-7}}\)
⇒ nI = 7961 ≈ 8000
Here; the product of n I is 8000. So,
Current I = 8A and no. of turns = 1000
The other design is I = 10A and n = 800/m. This is the most appropriate design as the requirement.

Question 3.
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from Its centre is given by,
B = \(\frac{\mu_0 \mathbf{I R}^2 \mathbf{N}}{2\left(\mathbf{x}^2+\mathbf{R}^2\right)^{3 / 2}}\)
a) Show that this reduces to the familiar result for field at the centre of the coil.
Solution:
Given, magnetic field at distance x :
B = \(\frac{\mu_0 \mathrm{NIR}^2}{2\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}}\) (∵ x = 0)
∴ The magnetic field at the centre B = \(\frac{\mu_0 \mathrm{~N} \mathrm{I} \mathrm{R}^2}{2 \mathrm{R}^3}\)
B = \(\frac{\mu_0 \mathrm{~N} \mathrm{I}}{2 \mathrm{R}}\)
This result is same as the magnetic field due to current loop at its centre.

b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated bya distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, B = 072\(\frac{\mu_0 \mathrm{NI}}{\mathrm{R}}\) approximately.
ISuch an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] :
Solution:
Radius of two parallel co-axial coils = R, number of turns = N Current = I
Let the mid points between the coils is at point O and P be the pomt around the mid point O.
Suppose, the distance between OP = d which is very less than R (d < < R)
For the Coil A,
O_AP = \(\frac{R}{2}\) + d
the magnetic field at point P due to coil A.


as according to the question d < <R, so neglect turn d².

The direction of B_A is along PO_B. According to Maxwell’s right hand rule
for the coil B, O_BP = (Rl₂ – d)
The magnetic field at point P due to coil B

The direction of magnetic field B_B is towards PO_B. So, the resultant magnetic field at P due to coil A and coil B is

Now, use binomial theorem and neglect higher power as d < < R

Question 4.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around whIch 3500 turns of a wire are wound. If the current in the wire Is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Solution:
a) For outside the toroid, the magnetic field is zero, because the magnetic field due to toroid is only inside it and along the length of toroid.

b) Inner radius of toroid, r₁ = 25 cm = 0.25 m
Outer radius of toroid, r₂ = 26 cm = 0.26 m
Number of turns N = 3500
current in the wire, I = 11A
The mean radius of toroid r = \(\left(\frac{\mathrm{r}_1+\mathrm{r}_2}{2}\right)\) = \(\frac{2}{2}\)(0.25 + 0.26) = 0.51
∴ length of toroid = 2πr = 2π × 0.51
the magnetic field strength due to toroid is B = µ₀ n I
where n is number of turns per unit length
n = \(\frac{N}{I}\)
B = 4π × 10^-7 × \(\frac{3500}{\pi \times 0.51}\) × 11 = 3.02 × 10^-2 T

c) The magnetic field in the empty space surrounded by toroid is also zero, because the
magnetic field due to toroid is only along its length.

Question 5.
Answer the following questions.
a) A magnetic field that varies in magnitude form point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
Answer:
The magnetic field is in constant direction from east to west. According to the question, a charged particle travels undeflected along a striaght path with constant speed. It is only possible, if the magnetic force experienced by the charged particle is zero. The magnitude of magnetic force on a moving charged particle in a magnetic field is given by F = qvB sin θ. Here F = 0, if and only if sinθ. This indicates the angle between the velocity and magnetic fled is 0° or 180°. Thus, the charged particle moves parallel or antiparallel to the magnetic field B.

b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory Would its final speed equal the initial speed lilt suffered no collisions with the environment?
Answer:
Yes, the final speed be equal to its initial speed as the magnetic force acting on the charged particle only changes the direction of velocity of charged particle but cannot change the magnitude of velocity of charged particle.

c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Solution:
B should be in a vertically downward direction.

Question 6.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Solution:
a) Since magnetic field is perpendicular to initial velocity of electron therefore, the electron will move in circular path.
B = 0.15 T,PD = 200 V
K. E. of electron = eV = \(\frac{1}{2}\)mv²
V² = \(\frac{2 \mathrm{ev}}{\mathrm{m}}\)
V = \(\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{9.1 \times 10^{-31}}}\)
V = \(\frac{8}{3}\) (e = charge of electron)
V = 2.66 × 10⁷ m/s
Also
BqV = \(\frac{\mathrm{m} \mathrm{V}^2}{\mathrm{Br}}\)
∴ r = \(\frac{\mathrm{mv}}{\mathrm{Bq}}\)
or r = \(\frac{9.1 \times 10^{-31} \times 2.66 \times 10^7}{0.15 \times 1.6 \times 10^{-19}}\)
= \(\frac{91 \times 83.86 \times 10^{-5}}{15 \times 1.6}\) = 10^-3 m = 1 mm

b) When magnetic field makes an angle 30° with the initial velocity i.e. θ = 30°
Then V¹ = V sin θ = \(\sqrt{\frac{2 \mathrm{ev}}{\mathrm{m}}}\) = sin 30°
V¹ = \(\frac{8}{3}\) × 10⁷ × \(\frac{1}{2}\) = \(\frac{4}{3}\) × 10⁷ m/s
The radius of the helical path is
r = \(\frac{\mathrm{m} \mathrm{V}^{\prime}}{\mathrm{Be}}\)
= \(\frac{9 \times 10^{-31} \times\left(\frac{4}{3} \times 10^7\right)}{0.15 \times 1.6 \times 10^{-19}}\) = 0.5 mm

Question 7.
A magnetic field set up using Helmholtz coils (described in Exercise 16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^-5 V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique ?
Solution:
B = 0.75 T, E = 9 × 10⁵ Vm^-1
V = \(\frac{E}{B}\) = \(\frac{9 \times 10^5}{0.75}\) = 12 × 10⁶ m/s
K. E of charged particle
= \(\frac{1}{2} \mathrm{mv}^2\) = eV
or \(\frac{\mathrm{e}}{\mathrm{m}}\) = \(\frac{v^2}{2 v}\) = \(\frac{144 \times 10^{12}}{2 \times 15000}\) = 4.8 × 10⁷C kg^-1
Particle is deuteron; the answer is not unique because only the ratio of charge to mass is determined. Other possibe answers are He⁺⁺, Li⁺⁺⁺ etc.
∴ He⁺⁺ and Li⁺⁺⁺ also have the same value of
e/m (∵ e/m = \(\frac{2 \mathrm{e}}{2 \mathrm{~m}}\) = \(\frac{3 \mathrm{e}}{3 \mathrm{~m}}\))

Question 8.
A straight horizontal conducting rod of length 0.45 m and mass 6 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
Solution:
Rod carrying current placed in uniform magnetic field expëriences a force Bu which is balanced by weigth of rod
F = B I l
mg = B I l
B = \(\frac{\mathrm{m} \mathrm{g}}{\mathrm{I} l}\) = \(\frac{60 \times 9.8}{1000 \times 0.45 \times 5}\) = 0.26 T

b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s^-2.
Solution:
Force due to magnetic field = B I l
= 0.26 × 5 × 0.45 = 5.85 N
Weight of rod = \(\frac{60}{1000}\) × 9.8 = 0.588 N
Total force = 0.588 + 0.585 = 1.173 N

Question 9.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Solution:
F = \(\frac{\mu_0}{4 \pi} \times \frac{2 I_1 I_2}{r}\)
= 10^-7 × \(\frac{2 \times 300 \times 300 \times 100}{1.5 \times 10^{-2}}\) = 1.2 Nm^-1 (Repulsive force)

Question 10.
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

a) the wire intersects the axis,
Solution:
F = BI 1 sin 900 = 1.5 × 7 × \(\frac{20}{100}\) = 2.1 N acting vertically downwards.

b) The wire is turned form N-S to north-east-northwest direction,
Solution:
Now θ = 45°, and length of the wire in the cylindrical region of the magnetic field is l, and is given by
l = l₁ sin 45°
l₁ = \(\frac{l}{\sin 45^{\circ}}\) = \(\sqrt{2 l}\)
So force F₁ = BI l sin 45°
= 1.5 × 7.0 × \(\sqrt{2} l\) × \(\frac{l}{\sqrt{2}}\)
= 10.5 × 0.2 = 2.1 N

c) The wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Solution:
When the wire is lowered through a distance 6.0 cm and reaches at CD then length of wire in magnetic field
l₂ = 2x
x.x = 4 × (10 + 6) = 64
x = 8 cm
∴ l₂ = 8 × 2 = 16 cm = \(\frac{16}{100}\)m
∴ Force on the wire
F₂ = BI l₂
= \(\frac{1.5 \times 7 \times 16}{100}\) = 1.68 N (Vertically downwards)

Question 11.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. ? What is the force on each case ? Which case corresponds to stable equilibrium?

Solution:
a) Torque = 300, G = 3000 × 10^-4 T
Area A = 10 × 5 = 50 cm² = \(\frac{50}{100 \times 100}\)m²
\(\tau\) = BIA = \(\frac{3}{10}\) × 12 × \(\frac{50}{100 \times 100}\) = 18 × 10^-2 Nm along Y-axis
b) Same as in (a)
c) 1.8 × 10^-2 Nm along x – direction
d) 1.8 × 10^-2 Nm at an angle 240° with the +x direction :
e) Zero
f) Zero
Force is zero in each case. Case (e) corresponds to stable and case (f) to unstable equilibrium.

Question 12.
A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. 1f the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^-5 m², and the free electron density in copper is given to be about 10²⁹ m^-3.)
Solution:
n = 20, r = 10 cm, B = 0.10 T, I = 50 A
a) Torque acting on the coil = nBIA sinθ = 0
b) Force acting on the coil = BIl sinθ = 0
e) Force on each electron
n = No. of electrons per unit volume
A = Area of cross – section of wire
F = Be V = \(\frac{\mathrm{BI}}{\mathrm{nA}}\) = \(\frac{0.1 \times 5}{10^{29} \times 10^{-5}}\) = 5 × 10^-25N

Question 13.
A solenoid 60 cm long and of radius 7.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two lead parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? g = 9.8 m s^-2.
Solution:
B = μ₀n I
Force acts on the wire normal to its length
∴ F = BI¹ l
= μ₀n I I¹ l and n = \(\frac{\frac{900}{60}}{100}\) = 1500
= μ₀n I I¹ l = mg
I = \(\frac{2.5 \times 9.8}{1000 \times 4 \pi \times 10^{-7} \times 1500 \times 6 \times \frac{1}{15}}\) = 108 A.

Question 14.
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA How will you convert the metre into a voltameter of range 0 to 18V ?
Solution:
Here G = 12Ω, Ig = 3mA = 3 × 10^-3A, V = 18V
R = \(\frac{v}{I_g}\) – G = \(\frac{18}{3 \times 10^{-3}}\) – 12 = (6000 – 12)Ω = 5988 Ω.

Question 15.
A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 A?
Solution:
Here G = 15Ω, I_g = 4 mA = 0.004 A, I = 6A
shunt S = \(\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\) = \(\frac{0.004 \times 15}{6-0.004}\) = \(\frac{0.06}{5.996}\) = ≈ 10 × 10^-3Ω = 10 m Ω
So shunt resistance = 10 m \(\simeq\) 10 mΩ

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(e) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(e)

I. Evaluate the following integrals.

Question 1.
∫\(\frac{x-1}{(x-2)(x-3)}\)dx
Solution:

Question 2.
∫\(\frac{x^2}{(x+1)(x+2)^2}\)dx
Solution:
∫\(\frac{x^2}{(x+1)(x+2)^2}\) ≡ \(\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
⇒ x² = A(x + 2)² + B(x + 1)(x + 2) + (x + 1) …………….. (1)
Put x = -2 in (1)
(-2)² = A(0) + B(0) + C(-2 + 1) ⇒ C = -4
Put x = -1 in (1)
(-1)² = A(-1 + 2)² + B(0) + C(0)
⇒ A = 1
Equation coeffs. of x² in (1)
1 = A + B
⇒ B = 1 – A = 1 – 1 = 0
∴ \(\frac{x^2}{(x+1)(x+2)^2}=\frac{1}{x+1}+\frac{0}{x+2}+\frac{(-4)}{(x+2)^2}\)
∴ ∫\(\frac{x^2}{(x+1)(x+2)^2}\)dx
= ∫\(\frac{1}{x+1}\)dx – 4∫\(\frac{1}{(x+2)^2}\)dx
= log|x + 1| – 4\(\frac{(-1)}{x+2}\)
= log|x + 1| + \(\frac{4}{x+2}\) + C

Question 3.
∫\(\frac{x+3}{(x-1)(x^2+1)}\)dx
Solution:
Let \(\frac{x+3}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\)
⇒ (x + 3) = A(x² + 1) + (Bx + C)(x – 1) …………….. (1)
Put x = 0 in (1)
3 = A(1) + C(-1)
⇒ A – C = 3 ⇒ C = A – 3 = 2 – 3 = -1
Equation coefficient of x² in (1)
0 = A + B
⇒ B = -A = -2
∴ \(\frac{x+3}{(x-1)(x^2+1)}=\frac{+2}{(x-1)}+\frac{-2x-1}{x^2+1}\)
∫\(\frac{x+3}{(x-1)(x^2+1)}\)dx = 2∫\(\frac{1}{x-1}\)dx
-∫\(\frac{2x}{x^2+1}\)dx – ∫\(\frac{1}{x^2+1}\)dx
= 2 log |x – 1| – log |x² + 1| – tan^-1(x) + C

Question 4.

Solution:

Question 5.
∫\(\frac{dx}{(e^x+e^{2x}}\)
Solution:

Question 6.
∫\(\frac{dx}{(x+1)(x+2)}\)
Solution:

Question 7.
∫\(\frac{1}{(e^x-1}\)dx
Solution:

Question 8.
∫\(\frac{1}{(1-x)(4+x^2)}\)dx
Solution:
Let \(\frac{1}{(1-x)(4+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{4+x^2}\)
⇒ 1 = A(4 + x²) + (Bx + C)(1 – x) ………….. (1)
Put x = 1 in (1)
1 = A(4 + 1) ⇒ A = \(\frac{1}{5}\)
Put x = 0 in (1)
1 = A(4) + C(1)
⇒ C = 1 – 4A = 1 – 4(\(\frac{1}{5}\)) = \(\frac{5-4}{5}\) = \(\frac{1}{5}\)
0 = A – B
⇒ B = A = \(\frac{1}{5}\)

Question 9.
∫\(\frac{2x+3}{x^3+x^2-2x}\)dx
Solution:

Put x = 0 in (1), then
3 = A(2)(-1) + B(0) + C(0)
⇒ A = –\(\frac{3}{2}\)
Put x = 1 in (1). Then
2 + 3 = A(0) + B(0) + C(1)(3)
⇒ C = \(\frac{5}{3}\)
Put x = -2 in (1). Then
2(-2) + 3 = A(0) + B(-2)(-2 – 1) + C(0)
⇒ -1 = 6B ⇒ B = \(\frac{-1}{6}\)

II. Evaluate the following integrals.

Question 1.
∫\(\frac{dx}{6x^2-5x+1}\)
Solution:

Question 2.
∫\(\frac{dx}{x(x+1)(x+2)}\)
Solution:
\(\frac{1}{x(x+1)(x+2)}\) ≡ \(\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}\)
⇒ 1 ≡ A(x + 1)(x + 2) + B(x)(x + 2) + C(x)(x + 1)
Put x = 0
1 = A(1)(2) + B(0) + C(0) ⇒ A = \(\frac{1}{2}\)
Put x = -1
1 = A(0) + B(0) + C(-2)(-2 + 1)
⇒ C = \(\frac{1}{2}\)

Question 3.
∫\(\frac{3x-2}{(x-1)(x+2)(x-3)}\)dx
Solution:
\(\frac{3x-2}{(x-1)(x+2)(x-3)}\) ≡ \(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
⇒ 3x – 2 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
Put x = 1
3(1) – 2 = A(1 + 2)(1 – 3) + B(0) + C(0)
⇒ A = \(\frac{-1}{6}\)
Put x = 3
3(3) – 2 = A(0) + B(0) + C(3 – 1)(3 + 2)
C = \(\frac{7}{10}\)
Put x = -2
3(-2) – 2 = A(0) + B(-2 – 1)(-2 – 3) + C(0) – 8
= 15B ⇒ B = \(\frac{-8}{15}\)

Question 4.
∫\(\frac{7x-4}{(x-1)^2(x+2)}\)dx
Solution:
\(\frac{7x-4}{(x-1)^2(x+2)}\) ≡ \(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\)
⇒ 7x – 4 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² ………….. (1)
Put x = 1 in (1)
7 – 4 = A(0) + B(1 + 2) ⇒ B = 1
Put x = -2 in (1)
7(-2) – 4 = A(0) + B(0) + C(-2 – 1)²
⇒ -18 = 9C ⇒ C = -2
Equating coeffs. of x² in (1)
0 = A + C ⇒ A = -C = 2

III. Evaluate the following integrals.

Question 1.
∫\(\frac{1}{(x-a)(x-b)(x-c)}\)dx
Solution:

Question 2.
∫\(\frac{2x+3}{(x+3)(x^2+4)}\)dx
Solution:
\(\frac{2x+3}{(x+3)(x^2+4)}\) = \(\frac{A}{x+3}+\frac{Bx+C}{x^2+4}\)
2x + 3 = A(x² + 4) + (Bx + C)(x + 3)
x = -3 ⇒ -3 = A(9 + 4) = 13A
A = –\(\frac{3}{13}\)
Equating the coefficient of x²
0 = A + B ⇒ B = -A = \(\frac{3}{13}\)
Equating the constants
3 = 4A + 3C

Question 3.
∫\(\frac{2x^2+x+1}{(x+3)(x-2)^2}\)dx
Solution:
Let \(\frac{2x^2+x+1}{(x+3)(x-2)^2}\) = \(\frac{A}{x+3}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\)
2x² + x + 1 = A(x – 2)² + B(x + 3)(x – 2) + C(x + 3)
x = 2 ⇒ 8 + 2 + 1 = C(2 + 3) = 5C
⇒ C = \(\frac{11}{5}\)
x = -3 ⇒ 18 – 3 + 1
= A(-5)² = 25 A ⇒ A = \(\frac{16}{25}\)
Equating the coefficients of x²
2 = A + B ⇒ B = 2 – A = 2 – \(\frac{16}{25}\) = \(\frac{34}{25}\)

Question 4.
∫\(\frac{dx}{x^3+1}\)dx
Solution:
\(\frac{1}{x^3+1}\) = \(\frac{1}{(x+1)(x^2-x+1)}\)
Let \(\frac{1}{x^3+1}\) = \(\frac{1}{x+1}+\frac{1}{x^2-x+1}\)
⇒ 1 = A(x² – x + 1) + (Bx + C)(x + 1) ……………. (1)
Put x = -1 in (1)
1 = A(1 + 1 + 1) + (-B + C)(0)
⇒ 3A = 1 ⇒ A = \(\frac{1}{3}\)
Put x = 0 in (1)
1 = A(1) + C(1)
⇒ C = 1 – A = 1 – \(\frac{1}{3}=\frac{2}{3}\)
Equating the coefficients of x²
O = A + B ⇒ B = -A = –\(\frac{1}{3}\)

Question 5.
∫\(\frac{\sin x \cos x}{\cos^2 x+3cos x+2}\)dx
Solution:
Put cos x = t ⇒ – sin x dx = dt
∫\(\frac{\sin x \cos x}{\cos^2 x+3cos x+2}\)dx = ∫\(\frac{-t dt}{t^2+3t+2}\)
= -∫\(\frac{t}{t^2+3t+2}\)dt …………. (1)
Let \(\frac{t}{t^2+3t+2}\) = \(\frac{t}{(t+1)(t+2)}\)
= \(\frac{A}{t+1}+\frac{B}{t+2}\)
⇒ t = A(t + 2) + B(t + 1) ………… (2)
Put t = -1 in (2)
-1 = A(-1 + 2) ⇒ A = -1
Put t = -2 in (2)
-2 = B(-2 + 1) ⇒ B = 2
∴ \(\frac{t}{t^2+3t+2}\) = \(\frac{-1}{t+1}+\frac{2}{t+2}\) ……….. (3)
∴ From (1) & (2)
∫\(\frac{\sin x.\cos x}{\cos^2 x+3cos x+2}\)dx
= -[∫\(\frac{-1}{t+1}\)dt+2∫\(\frac{1}{t+2}\)]
= ∫\(\frac{1}{t+1}\) – 2∫\(\frac{1}{t+2}\)
= log|t + 1| – 2log|t + 2| + C
= log|1 + cos x| – 2log|2 + cos x| + C
= log|1 + cos x| – log(2 + cos x)² + C
= log|\(\frac{1+\cos x}{(2+\cos x)^2}\)| + C

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(d)

I. Evaluate the following integrals.

Question 1.
∫\(\frac{dx}{\sqrt{2x-3x^2+1}}\)
Solution:

Question 2.
∫\(\frac{\sin \theta}{\sqrt{2-\cos^2 \theta}}\)dθ
Solution:

Question 3.
∫\(\frac{\cos x}{\sin^2 x+4sin x+5}\)dx
Solution:
t = sin x ⇒ dt = cos x dx
I = ∫\(\frac{dt}{t^2+4t+5}\) = ∫\(\frac{dt}{(t+2)^2+1}\)
= tan^-1(t + 2) + C
= tan^-1(sin x + 2) + C

Question 4.
∫\(\frac{dx}{1+\cos^2 x}\)
Solution:

Question 5.
∫\(\frac{dx}{2\sin^2 x+3\cos^2 x}\)
Solution:

Question 6.
∫\(\frac{1}{1+\tan x}\)dx
Solution:

Question 7.
∫\(\frac{1}{1-\cot x}\)dx
Solution:

II. Evaluate the following integrals.

Question 1.
∫\(\sqrt{1+3x-x^2}\)dx
Solution:

Question 2.
∫\(\frac{9\cos x-\sin x}{4\sin x+5\cos x}\)dx
Solution:

Question 3.
∫\(\frac{2\cos x+3\sin x}{4\cos x+5\sin x}\)dx
Solution:
Let 2 cos c + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x)
Equating the co-efficient of sin x and cos x,
we get
4A + 5B = 2
5A – 4B = 3

Question 4.
∫\(\frac{1}{1+\sin x+\cos x}\)dx
Solution:

Question 5.
∫\(\frac{1}{3x^2+x+1}\)dx
Solution:

Question 6.
∫\(\frac{dx}{\sqrt{5-2x^2+4x}}\)
Solution:

III. Evaluate the following integrals.

Question 1.
∫\(\frac{x+1}{\sqrt{x^2-x+1}}\)
Solution:

Question 2.
∫(6x + 5)\(\sqrt{6-2x^2+x}\)dx
Solution:
Let 6x + 5 = A(1 – 4x) + B
Equating the constants
A + B = 5
B = 5 – A = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)
∫(6x + 5)\(\sqrt{6-2x^2+x}\)dx

Question 3.
∫\(\frac{dx}{4+5\sin x}\)
Solution:
t = tan \(\frac{x}{2}\) ⇒ dt = sec² \(\frac{x}{2}\) . \(\frac{1}{2}\)dx

Question 4.
∫\(\frac{1}{2-3\cos 2x}\)dx
Solution:
t = tan x ⇒ dt = sec² x dx
= (1 + tan² x)dx
= (1 +t²)dx
dx = \(\frac{dt}{1+t^2}\)

Question 5.
∫x\(\sqrt{1+x-x^2}\)dx
Solution:
Let x = A(1 – 2x) + B
Equating the coefficients of x
1 = -2 A ⇒ A = –\(\frac{1}{2}\)
Equating the constants
0 = A + B ⇒ B = -A = \(\frac{1}{2}\)
∫x\(\sqrt{1+x-x^2}\)dx
= –\(\frac{1}{2}\)∫(1 – 2x)\(\sqrt{1+x-x^2}\)dx + \(\frac{1}{2}\)∫\(\sqrt{1+x-x^2}\)dx

Question 6.
∫\(\frac{dx}{(1+x)\sqrt{3+2x-x^2}}\)
Solution:

Question 7.
∫\(\frac{dx}{4\cos x+3\sin x}\)
Solution:
Let t = tan\(\frac{x}{2}\) so that dx = \(\frac{2dt}{1+t^2}\)

Question 8.
∫\(\frac{1}{\sin x+\sqrt{3} \cos x}\)dx
Solution:
Let t = tan \(\frac{x}{2}\) so that dx = \(\frac{2dt}{1+t^2}\)
sin x = \(\frac{2t}{1+t^2}\), cos x = \(\frac{1-t^2}{1+t^2}\)

Question 9.
∫\(\frac{dx}{5+4\cos 2x}\).
Solution:

Question 10.
∫\(\frac{2\sin x+3\cos x+4}{3\sin x+4\cos x+5}\)dx.
Solution:
Let 2 sin x + 3 cos x + 4
= A(3 sin x + 4 cos x + 5) + 3(3 cos x – 4 sin x) + C
Equating the co-efficient of
sin x, we get 3A – 4B = 2
cos x, we get 4A + 3B = 3

Equating the constants
4 = 5A + C
C = 4 – 5A = 4 – 5.\(\frac{18}{25}\) = \(\frac{2}{5}\)

Substituting in (1)
I = \(\frac{18}{25}\). x + \(\frac{1}{25}\) log|3 sin x + 4 cos x + 5| – \(\frac{4}{5\left(3+\tan \frac{x}{2}\right)}\) + C

Question 11.
∫\(\sqrt{\frac{5-x}{x-2}}\) dx on (2, 5).
Solution:

Let 5 – x = A. \(\frac{d}{dx}\)(7x – 10 – x²) + B
⇒ 5 – x = A(7 – 2x) + B
Equating coffs. of like terms
-2A = -1 ⇒ A = \(\frac{1}{2}\)
7A + B = 5
7(+\(\frac{1}{2}\)) + B = 5 ⇒ B = 5 – \(\frac{7}{2}\) = \(\frac{3}{2}\)
∴ 5 – x = \(\frac{1}{2}\)(7 – 2x) + \(\frac{3}{2}\)

Question 12.
∫\(\sqrt{\frac{1+x}{1-x}}\) dx on (-1, 1).
Solution:

Question 13.
∫\(\frac{dx}{(1 – x)\sqrt{3-2x+x^2}}\) on (-1, 3).
Solution:
Put 1 – x = \(\frac{1}{1}\) ⇒ 1 – \(\frac{1}{1}\) = x\(\frac{1}{1-x}\) = t
dx = \(\frac{1}{t^2}\)dt
3 – 2x – x² = 3 – 2(\(\frac{1-1}{1}\)) – (\(\frac{1-1}{1}\))²

Question 14.
∫\(\frac{dx}{(x + 2)\sqrt{x+1}}\) on (-1, ∞).
Solution:
Put = x + 1 = t² ⇒ dx = 2t dt and
x + 2 = 1 + t²

Question 15.
∫\(\frac{dx}{(2x + 3)\sqrt{x+2}}\) on I ⊂ (-2, ∞)\{\(\frac{-3}{2}\)}.
Solution:
Put x + 2 = t² ⇒ dx = 2t dt and
2x + 3 = 2(t² – 2) + 3 = 2t² – 1

Question 16.
∫\(\frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}}\)dx on (0, 1).
Solution:

Question 17.
∫\(\frac{dx}{(x + 1)\sqrt{2x^2+3x+1}}\) on
Solution:

Question 18.
∫\(\sqrt{e^x-4}\) dx on [log_e 4, ∞]
Solution:
Put e^x – 4 = t² ⇒ e^x dx = 2t dt

Question 19.
∫\(\sqrt{1+\sec x}\) dx on [(2n – \(\frac{1}{2}\))π – (2n + \(\frac{1}{2}\))π], (n ∈ Z).
Solution:

Question 20.
∫\(\frac{dx}{1+x^4}\) on R.
Solution:

Students get through AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system?
Answer:
Basic blocks in a communication system are

1. Transmitter
2. Receiver
3. Channel

Question 2.
What is the ‘World Wide Web” (WWW)?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to everyone round the clock throughout the year.

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

1. D Part of stratosphere (65-70 km day only),
2. E Part of stratosphere (100 km day only),
3. F₁ Part of mesosphere (170 km – 190 km),
4. F₂ Part of thermosphere [300 km at night 250 – 400 km during day time].

Question 6.
Define modulation. Why is it necessary ? [A.P. 17; A.P., T.S. Mar. 16, T.S. Mar. 15, Mar. 14]
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.
Necessary: Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. [T.S. Mar. 15, 17, A.P. Mar. 16]
Answer:
The basic methods of modulation are :

1. Amplitude modulation (AM)
2. Frequency modulation (FM)
3. Phase modulation (PM)

Question 8.
Which type of communication is employed in Mobile Phones ? [A.P. Mar. 15]
Answer:
Space wave mode of propagation is employed in mobile phones.

Textual Examples

Question 1.
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna Is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 × 10⁶ m.
Solution:
d_m = \(\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{Rh}_{\mathrm{R}}}\)
d_m = \(\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50 \mathrm{~m}}\)
=64 × 10² × \(\sqrt{10}\) +8 × 10³ × \(\sqrt{10}\)m = 144 × 10² × \(\sqrt{10}\)m
= 45.5 km.

Question 2.
A message signal of frequency 10 kHz and peak voltage of 10 volts Is used to modulate a carrier of frequency 1 MHz and peak voltage of 20 volts. Determine
(a) modulation index
(b) the side bands produced.
Solution:
a) Modulation index = \(\frac{10}{2}\) = 0.5
b) The side bands are at (1000 + 10 kHz) = 1010 kHz and (1000 – 10 kHz) = 990 kHz.

Students get through AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms

Very Short Answer Questions

Question 1.
What is the physical meaning of negative energy of an electron’ ?
Answer:
The ‘negative energy of an electron’ indicates that the electron is bound to the nucleus due to force of attraction.

Question 2.
Sharp lines are present in the spectrum of a gas. What does this indicate ?
Answer:
Sharp lines in the spectrum of gas, indicates bright lines against dark background.

Question 3.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Planck’s constant.

Question 4.
What is the difference between α – particle and helium atom ?
Answer:
Alpha particle

1. It is a + 2e charged Helium nucleus.
2. It contains 2 protons and 2 neutrons.

Helium atom

1. It has no charge.
2. It contains 2 protons, 2 electrons and 2 neutrons.

Question 5.
Among alpha, beta and gamma radiations, which get affected by the electric field ?
Answer:
Alpha and Beta radiations are get affected by the electric field.

Question 6.
What do you understand by the phrase ground state atom ?
Answer:
If the electron is present in the ground state, it is called ground state atom.

Question 7.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment ?
Answer:
The size of the atom is 10^-10 m and size of the nucleus is 10^-15 m. Hence atom has large empty space. So the mass of nucleus has no significance in Rutherford’s scattering experiment.

Question 8.
The Lyman series of hydrogen spectrum lies in the ultraviolet region. Why ? [A.P. Mar. 15]
Answer:
The calculated values of wavelengths lie in the ultraviolet region of the spectrum will agree with the values of wavelengths observed experimentally by Lyman.

Question 9.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:
Wavelength limits of some spectral series of hydrogen.

Question 10.
Give two drawbacks of Rutherford’s atomic model.
Answer:
Drawbacks of Rutherford’s atomic model:

1. As the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. But matter .is stable, we cannot expect the atom collapse.
2. The atoms should emit continuous spectrum, but what we observe is only a line spectrum.

Question 11.
If the kinetic energy of revolving electron in an orbit is K, what is its potential energy and total energy ?
Answer:
For an electron revolving round the nucleus, total energy is always negative and it is numerically equal to kinetic energy.
∴ Total energy = -Kinetic energy = -K
Potential energy is always negative and PE = 2 × TE = -2K

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering ? How are they related to each other ?
Answer:

1. Impact parameter (b) : Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
2. Scattering angle (θ) : The scattering angle (θ) is the angle between the asymtotic direction of approach of the α – particle and the asymtotic direction in which it receeds.
3. The relation between b and θ is b = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Ze}^2}{\mathrm{E}}\) cot \(\frac{\theta}{2}\) where E = K.E of α – particle = \(\frac{1}{2}\) mυ².

Question 2.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :

1. Suppose an α-particle with initial kinetic energy (K.E) is directed towards the centre of the nucleus of an atom.
2. On account of Coulomb’s repulsive force between nucleus and alpha particle, kinetic energy of alpha particle goes on decreasing and in turn, electric potential energy of the particle goes on increasing.
3. At certain distance d’ from the nucleus, K. E of α-particle reduces to zero. The particle stops and it cannot go closer to the nucleus. It is repelled by the nucleus and therefore it retraces its path, turning through 180°.
4. Therefore, the distance d is known as the distance of closest of approach.
   
   The closest distance of approach,
   d = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{Z e^2}{\left(\frac{1}{2} m v^2\right)}\)
5. Impact parameter (b) : Impact parameter is defined as the ⊥^r distance of the initial velocity vector of the α – particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
   

Question 3.
Describe Rutherford atom model. What are the draw backs of this model ?
Answer:
Rutherford atom model: The essential features of Rutherford’s nuclear model of the atom or planetary model of the atom are as follows :

1. Every atom consists of tiny central core, called the atomic nucleus, in which the entire positive charge and almost entire mass of the atom are concentrated.
2. The size of nucleus is of the order of 10^-15m, which is very small as compared to the size of the atom which is of the order of 10^-10 m.
3. The atomic nucleus is surrounded by certain number of electrons. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus is equal to total positive charge on the nucleus.
4. These electrons revolve around the nucleus in various circular orbits as do the planets around the sun. The centripetal force required by electron for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

Drawbacks : According to classical E.M. theory,

1. the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. As matter is stable, we cannot expect the atoms to collapse.
2. since the frequency of radiation emitted is the same as the frequency of revolution, the atom should radiate a continuous spectrum, but what we observe is only a line spectrum.

Question 4.
What are the limitations of Bohr’s theory of hydrogen atom ? [A.P. Mar. 17; Mar. 14]
Answer:
Limitations of Bohr’s theory of Hydrogen atom :

1. This theory is applicable only to simplest atom like hydrogen, with z = 1. The theory fails in case of atoms of other elements for which z > 1.
2. The theory does not explain why orbits of electrons are taken as circular, while elliptical orbits are also possible.
3. Bohr’s theory does not say anything about the relative intensities of spectral lines.
4. Bohr’s theory does not take into account the wave properties of electrons.

Question 5.
Write a short note on Debroglie’s explanation of Bohr’s second postulate of quantization. [T.S. Mar. 17]
Answer:
Debroglie’s explanation of Bohr’s second postulate of quantization :

1. The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized i.e., mυr = \(\frac{\mathrm{nh}}{2 \pi}\) where n = 1, 2, 3,….
2. According to Debroglie, the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave.
3. When a string fixed at two ends is plucked, a large number of wavelengths are excited and standing wave is formed.
4. It means that in a string, standing waves form when total distance travelled by a wave down the string and back is an integral number of wavelengths.
5. According to Debroglie, a stationary orbit is that which contains an integral number of Debrogile waves associated with the revolving electron.
6. For an electron revolving in n^th circular orbit of radius r_n, total distance covered = circumference of the orbit = 2πr_n
   ∴ For permissible orbit, 2πr_n = nλ
7. According to Debrogile, λ = \(\frac{h}{m v_n}\)
   Where υ_n is speed of electron revolving in n^th orbit
   ∴ 2πr_n = \(\frac{\mathrm{nh}}{\mathrm{m} v_{\mathrm{n}}}\)
   mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}=\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)
   i.e., angular momentum of electron revolving in n^th orbit must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\), which is the quantum condition proposed by Bohr in second postulate.

Question 6.
Explain the different types of spectral series in hydrogen atom. [T.S. Mar. 16, A.P. Mar. 15]
Answer:
The atomic hydrogen emits a line spectrum consisting of five series.

1. Lyman series : v = Rc \(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) where n = 2, 3, 4
2. Balmer series : v = Rc \(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) where n = 3, 4, 5.
3. Paschen series : v = Rc \(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\) where n = 4, 5, 6
4. Brackett series: v = Rc \(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) where n = 5, 6, 7
5. Pfund series: v = Rc \(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) where n = 6, 7, 8

Long Answer Questions

Question 1.
State the basic postulates of Bohr’s theory of atomic spectra. Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom. [A.P. Mar. 16]
Answer:
a) Basic postulates of Bohr’s theory are

1. The electron revolves round a nucleus in an atom in various orbits known as stationary orbits. The electrons cannot emit radiation when moving in their own stationary levels.
2. The electron can revolve round the nucleus only in allowed orbits whose angular momentum is the integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)
   i.e., mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}\) …………… (1)
   where n = 1, 2, 3 …………..
   If an electron jumps from higher energy (E₂) orbit to the lower energy (E₁) orbit, the difference of energy is radiated in the form of radiation.
   i.e., E = hv = E₂ – E₁ ⇒ v = \(\frac{E_2-E_1}{h}\) …………… (2)

b) Energy of emitted radiation: In hydrogen atom, a single electron of charge – e, revolves around the nucleus of charge e in a circular orbit of radius r_n.
1) K.E. of electron : For the electron to be in circular orbit, centripetal force = The electrostatic force of attraction between the electron and nucleus.

3) Radius of the oribit: Substituting the value of (6) in (2),
\(\frac{\mathrm{m}}{\mathrm{r}_{\mathrm{n}}}\left(\frac{\mathrm{n}^2 \mathrm{~h}^2}{4 \pi^2 \mathrm{r}_{\mathrm{n}}^2 \mathrm{~m}^2}\right)=\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}^2}\)
r_n = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\) ……………. (1)
∴ r_n = 0.53 n²

4) Total energy (E_n) : Revolving electron possess K.E. as well as P.E.

Problems

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 x HTum. What is the radius of the second orbit ?
Solution:
∴ r_n ∝ n²
\(\) ⇒ r₂ = 4r₁
∴ r₂ = 4 × 5.3 × 10^-11 = 2.12 × 10^-10m.

Question 2.
The total energy of an electron in the first excited state of the hydrogen atom is -3.4eV. What is the potential energy of the electron in this state?
Solution:
In 1^st orbit, E = -3.4 eV
Total energy E = \(\frac{\mathrm{KZe}^2}{2 \mathrm{r}}-\frac{\mathrm{KZe}^2}{\mathrm{r}}\)
\(\frac{\mathrm{KZe}^2}{\mathrm{r}}\) = U(say)
E = \(\frac{\mathrm{U}}{2}\) – u = \(\frac{\mathrm{-U}}{2}\)
U = -2E
∴ U = -2 × -3.4 = 6.8 eV.

Question 3.
The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ?
Solution:
In Hydrogen like atom, we know that
K = – Total energy E
Here E = – 3.4eV
∴ K = -(-3.4) = 3.4 eV . . ,

Question 4.
Prove that the ionisation energy of hydrogen atom is 13.6 eV.
Solution:
n = 1 corresponds to ground state.
E = \(\frac{-13.6}{\mathrm{n}^2}\) eV
E = \(\frac{-13.6}{\mathrm{1}^2}\) eV
E = -13.6 eV
The minimum energy required to free the electron from the ground state of hydrogen atom = 13.6 eV.
∴ Ionisation energy of hydrogen atom = 13.6 eV

Question 5.
Calculate the ionization energy for a lithium atom.
Solution:
For 3^Li7 atom, Z = 3, n = 2 [∵Li = 1s² 2s¹]
E_n = \(\frac{13.6 \mathrm{Z}^2}{\mathrm{n}^2}\) eV
= \(\frac{13.6 \times(3)^2}{4}\) = 30.6 eV
∴ Ionization energy of Lithium = 30.6eV

Question 6.
The wavelength of the first member of Lyman series is 1216 A. Calculate the wavelength of second member of Balmer series.
Solution:

Textual Examples

Question 1.
In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10^-15m) is analogous to the sun about which the electron move in orbit (radius ≈ 10^-10m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.5 × 10¹¹m. The radius of sun is taken as 7 × 10⁸ m.
Solution:
The ratio of the radius of electron’s orbit to the radius of nucleus is (10^-10 m)/(10^-15 m) = 105, that is, the radius of the electron’s orbit is 10⁵ times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 10⁵ times larger than the radius of the sun, the radius of the earth’s orbit would be 10⁵ × 7 × 10⁸ m = 7 × 10¹³ m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun.

It implies that an atom contains a much greater fraction of empty space than our solar system does.

Question 2.
In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV α-particle before it comes momentarily to rest and reverses its direction?
Solution:
The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. The system’s initial mechanical energy is E_i; before the particle and nucleus interact, and it is equal to its mechanical energy E_f when the α-particle momentarily stops. The initial energy E_t is just the kinetic energy K of the incoming α-particle. The final energy E_f is just the electric potential energy U of the system. The potential energy U can be calculated from Equation.
F = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}^2}\)
Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy
E_i = E_f as
K = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{d}}=\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~d}}\)
Thus the distance of closest approach d is given by
d = \(\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~K}}\)
The maximum kinetic energy found in α-particles of natural origin is 7.7 MeV or 1.2 × 10^-12 J. Since 1/4πε₀ = 9.0 × 10⁹N m²/C². Therefore with e = 1.6 × 10^-19C, we have,
d = \(\frac{(2)\left(9.0 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2 \mathrm{Z}}{1.2 \times 10^{-12} \mathrm{~J}}\)
= 3.84 × 10^-16 Zm
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3.0 × 10^-14m = 30 fm. (1 fm (i.e. fermi) = 10^-15m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10^-14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the a-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus.

Question 3.
It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
Solution:
Total energy of the electron in hydrogen atom is – 13.6 eV = -13.6 × 1.6 × 10^-19J
= -2.2 × 10^-18 J.
Thus from equation, E = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) we have
– \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = 2.2 10^-18 J
This gives the orbital radius
r = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^-11 m.
The velocity of the revolving electron can be computed from Equation r = –\(\frac{e^2}{4 \pi \varepsilon_0 r m v^2}\)
with
m = 9.1 × 10^-31kg,
υ = \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{mr}}}\) = 2.2 × 10⁶ m/s

Question 4.
According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Solution:
From Examjple 3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^-11 m is 2.2 × 10^-6 m/s. Thus, the frequency of the electron moving around the proton is
v = \(\frac{v}{2 \pi \mathrm{r}}=\frac{2.2 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
≈ 6.6 × 10¹⁵ Hz.
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵Hz.

Question 5.
A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.
Solution:
From equation, we have
mυ_nr_n = nh/2π
Here m = 10 kg and r_n = 8 × 10⁶ m. We have the time period T of the circling satellite as 2h. That is T = 7200 s.
Thus the velocity υ_n = 2π r_n/T
The quantum number of the orbit of satellite
n = (2π r_n)² × m(T × h)
Substituting the values,
n = (2π × 8 × 10⁶m)² × 10/(7200 s × 6.64 × 10^-34 J s)
= 5.3 × 10⁴⁵
Note that the quantum number for the satellite motion is extremely large. In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.

Question 6.
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Solution:
The Rydberg formula is
hc/λ_if = \(\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right)\)
The wavelengths of the first four lines in the Lyman series correspond to transitions from n_i = 2, 3, 4, 5 to n_f = 1. We know that
\(\frac{\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}\) = 13.6 eV = 21.76 × 10^-19 J
Therefore,

Substituting, n_i = 2, 3, 4, 5 we get λ₂₁ = 1218 Å, λ₃₁ = 1028 Å, λ₄₁ = 974.3 Å, and λ₅₁ = 951.4 Å.

Students get through AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays”? [A.P. Mar. 17]
Answer:
Cathode rays are streams of fast-moving electrons or negatively charged particles.

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan’s experiment established that electric charge is quantized. That means the charge on anybody (oil drop) is always an integral multiple of the charge of an electron, i.e., Q = ne.

Question 3.
What is “work function”? [T.S. Mar. 17, 15]
Answer:
The minimum energy required to liberate an electron from a photo metal surface is called the work function, Φ₀.

Question 4.
What is “photoelectric effect” ?
Answer:
When light of sufficierit energy is incident on the photometal surface, electrons are emitted. This phenomenon is called photoelectric effect.

Question 5.
Give examples of “photosensitive substances”. Why are they called so ?
Answer:
Examples of photosensitive substances are Li, Na, K, Rb and Cs etc.
The work function of alkali metals is very low. Even the ordinary visible light, alkali metals can produce photoelectric emission. Hence they are called photosensitive substances.

Question 6.
Write down Einstein’s photoelectric equation. [A.P. Mar. 15]
Answer:
Einstein’s photoelectric equation is given by K_max = \(\frac{1}{2}\) mv²_max = hυ – Φ₀

Question 7.
Write down de-Broglie’s relation and explain the terms therein. [T.S. & A.P. Mar. 16]
Answer:
The de-Broglie wave length (λ) associated with a moving particle is related to its momentum (p) is λ = \(\frac{h}{p}=\frac{h}{m v}\), where h is planck’s constant.

Question 8.
State Heisenberg’s Uncertainity Principle. [Mar. 14]
Answer:
Uncertainity principle states that “it is impossible to measure both position (∆x) and momentum of an electron (∆p) [or any other particle] at the same time exactly”, i.e., ∆x . ∆p ≈ h where ∆x is uncertainty in the specification of position and ∆p is uncertainty in the specification of momentum.

Question 9.
The photoelectric cut off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ? [Mar. 11]
Answer:
Cut off voltage, V₀ = 1.5 V; Maximum kinetic energy, (KE)_max = eV₀ = e × 1.5 = 1.5eV

Question 10.
An electron, an α-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Answer:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p²/2m
Then, λ = h/\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses.A proton is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton.
Hence, α-particle has the shortest de Broglie wavelength.

Qeustion 11.
What is-the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts ? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelngth λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Question 12.
If the wave length of electro magnetic radiation is doubled, what happens to energy of photon ? [IPE 2015 (TS)]
Answer:
E = \(\frac{\mathrm{hc}}{\lambda}\) ⇒ E ∝ \(\frac{1}{\lambda}\), since wave length of photon is doubled, its energy becomes halved.

Question 13.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10^-15 V s. Calculate the value of Planck’s constant.
Solution:
Given, slope of graph tan θ = 4.12 × 10^-15 V – s;

For slope of graph, tan θ = \(\frac{\mathrm{V}}{\mathrm{v}}\)
We know that hv = eV
\(\frac{\mathrm{V}}{\mathrm{v}}=\frac{\mathrm{h}}{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{e}}\) = 4.12 × 10^-15; h = 4.12 × 10^-15 × 1.6 × 10^-19 = 6.592 × 10^-34 J – s.

Question 14.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given, P = 1.388 × 10³ W/m²; λ = 550 nm = 550 × 10^-9 m ,
h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s
Energy of each photon E = \(\frac{\text { hc }}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10^-19 J
No. of photons incident on the earth’s surface, N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{1.388 \times 10^3}{3.66 \times 10^{-19}}\)
∴ N = 3.838 × 10²¹ photons/m² – s.

Question 15.
Show that the wavelength of electromagnetic radiation is equal to the de-Brogile wavelength of its quantum (photon). [Mar. 14]
Answer:
Wave length of electromagnetic wave of frequency v and velocity C is given by,
λ = \(\frac{\mathrm{C}}{\mathrm{v}} \Rightarrow \lambda=\frac{\mathrm{C}}{\mathrm{v}} \times \frac{\mathrm{h}}{\mathrm{h}}=\frac{\mathrm{h}}{\left(\frac{\mathrm{hv}}{\mathrm{C}}\right)}=\frac{\mathrm{h}}{\mathrm{p}}\)    (∵ \(\frac{\mathrm{hv}}{\mathrm{C}}\) = p)
Hence, we can say wavelength of electromagnetic radiation is equal to the de-Brogile wavelength.

Sample Problem :

Question 1.
Calculate the (a) momentum and (b) dE-Brogile wavelength of the electrons accelerated through a potential difference of 56 V. [Mar. 14]
Answer:
a) Mass of the electron, m = 9 × 10^-31 kg;
Potential difference, V = 56V
Momentum of electron, mv = \(\sqrt{2 \mathrm{eVm}}=\sqrt{2 \times\left(1.6 \times 10^{-19}\right) \times 56 \times 9 \times 10^{-31}}\) = 4.02 × 10^-24kg ms^-1

b) de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.62 \times 10^{-34}}{4.04 \times 10^{-24}}\) = 1.64 × 10^-10m

Short Answer Questions

Question 1.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of the effect of frequency of incident radiation on stopping potential:

1. The experimental set up is shown in fig.
   
2. Monochromatic light of sufficient energy (E = hv) from source ‘s’ is incident on photosensitive plate ‘C’ (emitter), electrons are emitted by it.
3. The electrons are collected by the plate A (collector), by the electric field created by the battery.
4. The polarity of the plates C and A can be reversed by a commutator.
5. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photo current stops or becomes zero is called stopping potential.
6. The experiment is repeated with different frequencies, and their different stopping potential are measured with voltmeter.
7. From graph, we note that
   
   • The values of stopping potentials are different for different frequencies.
   • The value of stopping potential is more negative for radiation of highef incident frequency.
   • The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.

Question 2.
What is the deBroglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ms^-1? What can we infer from this result ?
Answer:
Given, m = 0.12 kg; υ = 20 m/s; h = 6.63 × 10^-34 J-s;
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{0.12 \times 20}=\frac{6.63 \times 10^{-34}}{2.4}\)
∴ λ = 2.762 × 10^-34 m = 2762 × 10^-21 Å.
The wave length of ball is very very small. Hence, its motion can be observed.

Question 3.
What is the effect of (i) intensity of light (ii) potential on photoelectric current ?
Answer:
(i) Effect of intensity of light on photoelectric current:
1) When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted increases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.

ii) The effect of potential on photoelectric current:

1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
   
2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential.
3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

Sample Problem :

Question 1.
The work function of caesium metal is 2.14 eV When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential and (c) maximum speed of the emitted photoelectrons ? [A.P. Mar. 16]
Solution:
Given, Φ₀ = 2.14 eV; v = 6 × 10¹⁴ Hz
a) KE_max = hv – Φ₀ = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) – 2.14
∴ KE_max = 0.35 eV

b) KE_max = eV₀ ⇒ 0.35 eV = eV₀
∴ V₀ = 0.35 V

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation explain the effect of intensity and potential on photoelectric current ? How did this equation account for the effect of frequency of ‘ incident light on stopping potential ?
Answer:

1. Einstein postulated that a beam of light consists of small energy packets called photons or quanta.
2. The energy of photon is E = hv. Where ‘h1 is Planck’s constant; v is frequency of incident light (or radiation).
3. If the absorbed energy of photon is greater than the work function (Φ₀ = hυ₀), the electron is emitted with maximum kinetic energy i.e., k_max = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = eV₀ = hv – Φ₀. This equation is known as Einstein’s photoelectric equation.
4. Effect of intensity of light on photoelectric current:
   When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted decreases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.
   
5. The effect of potential on photoelectric current:
   • On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
     
   • On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential (v₀).
   • Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.
6. The effect of frequency of incident radiation on stopping potential:
   On increasing the frequency of incident light, the value of stopping potential goes on increasing gradually as shown in fig. That means k_max increases eV₀ also increases.
   
7. From the graph, we note that
   • For a given photosensitive metal, the cut off potential (v₀) varies linearly with the frequency of the incident radiation.
   • For a given photosensitive metal, there is a certain minimum cut off frequency υ₀ (called threshold frequency) for which the stopping potential is zero.
     
8. From the graph we note that
   • The value of cut-off potential is different for radiation of different frequency.
   • The value of stopping potential is more negative for radiation of higher incident frequency.
9.  From above experiments, it is found that, if the incident radiation is of higher frequency than that of threshold frequency, the photoelectric emission is. possible.

Textual Examples

Question 1.
Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 × 10^-3 W. (a) What is the energy of a photon in the light beam ? (b) How many photons per second, on an average, are emitted by the source ?
Solution:
a) Each photon has an energy
E = hv = (6.63 × 10^-34 J s) (6.0 × 10¹⁴ Hz)
= 3.98 × 10^-19 J

b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = NE.
Then N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{2.0 \times 10^{-3} \mathrm{~W}}{3.98 \times 10^{-19} \mathrm{~J}}\)
= 5.0 × 10¹⁵ photons per second.

Question 2.
The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. [A.P. Mar. 16]
Solution:
a) For the cut-off or threshold frequency, the energy hv₀ of the incident radiation must be equal to work function Φ₀, so that
v₀ = \(\frac{\phi_0}{\mathrm{~h}}=\frac{2.14 \mathrm{eV}}{6.63 \times 10^{-34} \mathrm{Js}}\)
= \(\frac{2.14 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\) = 5.16 × 10¹⁴ Hz
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.

b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V₀ by the retarding potential V₀. Einstein’s Photoelectric equation is
eV₀ = hv – Φ₀ = \(\frac{\mathrm{hc}}{\lambda}\) – Φ₀
or λ = hc/(eV₀ + Φ₀)
= \(\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{(0.60 \mathrm{eV}+2.14 \mathrm{eV})}\)
= \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{(2.74 \mathrm{eV})}\)
λ = \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{2.74 \times 1.6 \times 10^{-19} \mathrm{~J}}\) = 454 nm

Question 3.
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.
(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum ? (Take h = 6.63 × 10^-34 J s and 1 eV = 1.6 × 10^-19 J.)
(b) From which of the photosensitive materials with work functions listed in table and using the results of (i), (ii) and (iii) of (a) can you build a photoelectric device that operates with visible light ?

Solution:
a) Energy of the incident photon,
E = hv = hc/λ
E = (6.63 × 10^-34 J s) (3 × 10⁸ m/s)/λ.
= \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{\lambda}\)
i) For violet light,
λ₁ = 390 nm (lower wavelength end)
Incident photon energy,
E₁ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{390 \times 10^{-9} \mathrm{~m}}\)
5.10 × 10^-19 J = \(\frac{5.10 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
= 3.19 eV

ii) For yellow-green light,
λ₂ = 550 nm (average wavelength)
Incident photon energy,
E₂ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{550 \times 10^{-9} \mathrm{~m}}\)
= 3.62 × 10^-19 J = 2.26 eV.

iii) For red light,
λ₃ = 760 nm (higher wavelength end)
Incident photon energy,
E₃ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{760 \times 10^{-9} \mathrm{~m}}\)
= 2.62 × 10^-19 J = 1.64 eV

b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function Φ₀ of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with Φ₀ = 2.75 eV), K (with Φ₀ = 2.30 eV) and Cs (with Φ₀ = 2.14 eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with Φ₀ = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials.

Question 4.
What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4 × 10⁶ m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s ?
Solution:
a) For the electron :
Mass m= 9.11 × 10^-31 kg, speed υ = 5.4 × 10⁶ m/s. Then, momentum
P = mυ = 9.11 × 10^-31 (kg) × 5.4 × 10⁶ (m/s)
P = 4.92 × 10^-24 kg m/s
de Broglie wavelength, λ = h/p
= \(\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}\)
λ = 0.135 nm

b) For the ball:
Mass m’ = 0.150 kg, speed υ’ = 30.0 m/s.
Then momentum
p’ = m’υ’ = 0.150 (kg) × 30.0 (m/s)
p’ = 4.50 kg m/s
de Broglie wavelength λ’ = h/p’.
= \(\frac{6.63 \times 10^{\pm 34} \mathrm{Js}}{4.50 \times \mathrm{kg} \mathrm{m} / \mathrm{s}}\)
λ’ = 1.47 × 10^-34 m
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10^-19 times the size of the proton, quite beyond experimental measurement.

Question 5.
An electron, an a-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Solution:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p²/2m
Then, λ = h /\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton \(\left({ }_1^1 \mathrm{He}\right)\) is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton
Hence, α-particle has the shortest de Broglie wavelength.

Question 6.
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10^-4. Calculate the particle’s mass and identify the particle.
Solution:
de Broglie wavelength of a moving particle, having mass m and velocity υ :
λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}\)
Mass, m = h/A.
For an electron, mass m_e = h/λ_e υ_e
Now, we have υ/υ_e = 3 and
λ/λ_e = 1.813 × 10^-4
Then, mass of the particle,
m = m_e \(\left(\frac{\lambda_{\mathrm{e}}}{\lambda}\right)\left(\frac{v_{\mathrm{e}}}{v}\right)\)
m = (9.11 × 10^-31 kg) × (1/3) × (1/1.813 × 10^-4)
m = 1.675 × 10^-27 kg.
Thus, the particle, with this mass could be a proton or a neutron.

Question 7.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelength λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Students get through AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves

Very Short Answer Questions

Question 1.
Give anyone use of infrared rays. [T.S. Mar. 17; A.P. Mar. 16]
Answer:

1. Infrared radiation plays an important role in maintaining the Earth warm.
2. Infrared lamps are used in physical therapy.
3. Infrared detectors are used on Earth Satellites.
4. These are used in taking photographs during conditions of fog, smoke, etc.

Question 2.
How are infrared rays produced? How they can be detected?
Answer:
Vibrations of atoms and molecules can produce infrared rays. These waves can be detected by Thermopile, Bolometer, and IR photographic film.

Question 3.
How are radio waves produced? How can they detect it?
Answer:
Radio waves can be produced by rapid acceleration and deceleration of electrons in aerials (conductors). These can be detected by receivers of aerials.

Question 4.
If the wave length of E.M radiation is doubled, what happens to the energy of photon ? [IPE 2016 (TS)]
Answer:
If the wave length of electromagnetic radiation is doubled, then energy will be halved because energy is inversely proportional to.wavelength of electromagnetic waves.
E = hυ = hc/λ ⇒ E ∝ 1/λ (∵ hc is a constant)

Question 5.
What is the principle of production of electromagnetic waves ?
Answer:
If the charge is accelerated both the magnetic field and electric field will change with Space and time, then electromagnetic waves are produced.

Question 6.
What is the ratio of speed of infrared rays and ultraviolet rays in vaccum ?
Answer:
The ratio of speed of infrared rays and ultraviolet rays in vacuum is 1 : 1.
All electromagnetic waves travel with same speed 3 × 10⁸ m /s in vaccum.

Question 7.
What is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave ?
Answer:
E₀ = CB₀
Where E₀ = Amplitude of electric field.
B₀ = Amplitude of magnetic field.
C = velocity of light.

Question 8.
What are the applications of microwaves ? [A.P. Mar. 17; T.S. Mar. 15]
Answer:

1. Microwaves are used in Radars.
2. Microwaves are used for cooking purposes.
3. A radar using microwave can help in detecting the speed of automobile while in motion.

Question 9.
Microwaves are used in Radars, why ? [Mar. 14]
Answer:
As microwaves are of smaller wavelengths, hence they can be transmitted as a beam signal in a particular direction. Microwaves do not bend around the comers of any obstacle coming in their path.

Question 10.
Give two uses of infrared rays.
Answer:

1. Infrared rays are used for producing dehydrated fruits.
2. They are used in the secret writings on the ancient walls.
3. They are used in green houses to keep the plants warm.

Question 11.
How are microwaves produced ? How can they detected ? [A.P. Mar. 16; IPE 15]
Answer:
Microwaves can be produced using Klystron valve or Magnetrons.
Microwaves can be detected using point contact diodes.

Question 12.
The charging current for a capacitor is 0.6 A. What is the displacement current across its plates ?
Answer:
i = charging current for a capacitor = 0.6 A
i = i_d = ε₀ = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
∴ i = i_d = 0.6 A
∴ Displacement current (i_d) = 0.6 A.

Question 13.
What physical quantity is the same for X-rays of wavelength 10^-10m, red light of wavelength 6800 Å and radiowaves of wavelength 500in ?
Answer:
The speed in vaccum is same for all the given wavelengths, which is 3 × 10⁸ m/s.

Question 14.
A radio can tune into any station in the 7.5 MHz to 12MHz band. What is the corresponding wavelength band ?
Answer:
λ₁ = \(\frac{3 \times 10^8}{7.5 \times 10^6}\) = 40 m
λ₂ = \(\frac{3 \times 10^8}{12 \times 10^6}\) = 25 m
Thus wavelength band is 40m to 25m.

Question 15.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vaccum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave ?
Answer:
Here, B₀ = 510nT = 510 × 10^-9T
E₀ = CB₀ = 3 × 10⁸ × 510 × 10^-9 = 153 NC^-1.

Question 16.
Define displacement current
Answer:
Displacement current (I_d) is equal to ε₀ times to the rate of change of electric flux. Displacement current is not the current produced due to charge carried. But it is due to varying electric flux. It is the current in the sense that it produces a magnetic field.
I_d = ε₀ \(\frac{\mathrm{d} \phi_{\varepsilon}}{\mathrm{dt}}\)

Short Answer Questions

Question 1.
State six characteristics of electromagnetic waves.
Answer:
Characteristics of electromagnetic waves :

1. Electromagnetic waves are produced by accelerated charges.
2. Electromagnetic waves are transverse in nature.
3. Electromagnetic waves donot require material medium for their propagation.
4. Electromagnetic waves obey principle of superposition of waves.
5. Velocity of E.M waves in vaccum depends on permittivity and permeability of free space.
6. Electromagnetic waves carry energy and momentum.
7. Electromagnetic waves exert pressure when they strike a surface.

Question 2.
What is Greenhouse effect and its contribution towards the surface temperature of earth ?
Answer:
Green house effect: Temperature of the earth increases due to the radiation emitted by
the earth is trapped by atmospheric gases like CO₂, CH₄, N₂, chlorofluoro carbons etc., is called green house effect.

1. Radiation from the sun enters the atmosphere and heat the objects on the earth. These heated objects emit infrared rays.
2. These rays are reflected back to Earth’s surface and trapped in the Earth’s atmosphere. Due to this temperature of the earth increases.
3. The layers of carbon dioxide (CO₂) and low lying clouds prevent infrared rays to escape Earth’s atmosphere.
4. Since day-by-day the amount of carbondioxide in the atmosphere increases, more • infrared rays are entrapped in the atmosphere.
5. Hence the temperature of the Earth’s surface increases day by day.

Problems

Question 1.
A plane electromagnetic wave travels in vaccum along z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz. What is its wavelength ?
Answer:
Electric and magnetic fields \(\overline{\mathrm{E}}\) and \(\overline{\mathrm{B}}\) of an electromagnetic wave must be perpendicular to the propagation of electromagnetic wave. Hence they lie in X – Y plane mutually perpendicular to each other.
Frequency of wave, v = 30MHz = 30 × 10⁶Hz.; Velocity of light, C = 3 × 10⁸m/s
Wavelength of the wave, λ = \(\frac{C}{V}=\frac{3 \times 10^8}{30 \times 10^6}\) = 10m

Question 2.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator ?
Answer:
According to Maxwell, a charged particle oscillating with a frequency produces electro-magnetic waves of same frequency. Hence frequency of EM waves produced is, 10⁹Hz.

Textual Examples

Question 1.
A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, It is connected for charging in series with a resistor R = 1 M Q across a 2V battery (fig)- Calculate the magnetic field at a point P halfway between the centre and the periphery of the plates, after t = 10^-3 s. (The charge on the capacitor at time τ is q (t) = CV [1 – exp (-t/τ)], where the time constant τ is equal to CR.).
Solution:
The time constant of the CR circuit is τ = CR = 10^-3s. Then, we have
q(t) = CV [1 – exp (-t/τ) ]
= 2 × 10^-9 [1 – exp (-t /10^-3)
The electric field in between the plates at time t is
E = \(\frac{q(t)}{\varepsilon_0 A}=\frac{q}{\pi \varepsilon_0}\); A = π (1)² m² = area of the plates.

Consider now a circular loop of radius (1/2)m parallel to the plates passing through R The magnetic field B at all points on the loop is along the loop arid of the same value.
The flux Φ_E through this loop is
The flux Φ_E = E × area of the loop
= E × π × (\(\frac{1}{2}\))² = \(\frac{\pi \mathrm{E}}{4}=\frac{\mathrm{q}}{4 \varepsilon_0}\)
The displacement current
i_d = e₀ \(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\) = \(\frac{1}{4} \frac{\mathrm{dq}}{\mathrm{dt}}\) = 0.5 × 10^-6 exp (-1)
at t = 10^-3s. Now, applying Ampere-Maxwell law to the loop, we get
B × 2π × (\(\frac{1}{2}\)) = m₀.(i_c + i_d) = m₀(0 + i_d) = 0.5 × 10^-6 m₀ exp(-1)
or, B = 0.74 × 10^-13 T.

Question 2.
A plane electromagnetic wave of frequency 25 MHz travels in free space along the x – direction. At a particular point in space and time, E = \(6.3 \hat{\mathbf{j}}\) V/m. What is B at the point ?
Solution:
Using Eq. B₀ = [E₀/c] the magnitude of B is
B = \(\frac{\mathrm{E}}{\mathrm{c}}\)
= \(\frac{6.3 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 2.1 × 10^-8 T
To find the direction, we note that E is along y-direction and the wave propagates along x- axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, \((+\hat{\mathrm{j}})\) × \((+\hat{\mathrm{k}})\) = i, B is along the z-direction. Thus. B = 2.1 × 10^-8 \(\hat{\mathrm{k}}\) T

Question 3.
The magnetic field in a plane electromagnetic wave is given by
B_y = 2 × 10^-7 sin (0.5 × 10³ × + 1.5 × 10¹¹ t)T.
a) What is the wavelength and frequency of the wave ?
b) Write an expression for the electric field.
Solution:
a) Comparing the given equation with
B_y = B₀ sin \(\left[2 \pi\left(\frac{\mathrm{x}}{\lambda}+\frac{\mathrm{t}}{\mathrm{T}}\right)\right]\)
We get, λ = \(\frac{2 \pi}{0.5 \times 10^3}\) m = 1.26 cm,
and \(\frac{1}{\mathrm{~T}}\) = v = (1.5 × 10¹¹)/2π = 23.9 GHz

b) E₀ = B₀C = 2 × 10^-7 T × 3 × 10⁸ m/s = 6 × 10¹ V/m
The electric field component is perpendicular to the direction of propagation and the di-rection of magnetic field. Therefore, the electric field component along the z-axis is obtained as E_z = 60 sin (0.5 × 10³x + 1.5 × 10¹¹t) V/m.

Question 4.
Light with an energy flux of 18 W/cm² falls on a non reflecting surface at normal incidence. If the surface has an area of 20 cm² find the average force exerted on the surface during a 30 minute time span.
Solution:
The total energy falling on the surface is
U = (18 W/cm²) × (20 cm²) × (30 × 60)
= 6.48 × 10⁵ J
Therefore, the total momentum delivered (for complete absorption) is
P = \(\frac{\mathrm{U}}{\mathrm{C}}=\frac{6.48 \times 10^5 \mathrm{I}}{3 \times 10^8 \mathrm{m} / \mathrm{s}}\) = 2.16 × 10^-3 kg m/s
The average force exerted on the surface is
F = \(\frac{\mathrm{p}}{\mathrm{t}}=\frac{2.16 \times 10^{-3}}{0.18 \times 10^4}\) = 1.2 × 10^-6 N

Question 5.
Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3m. Assume that the efficiency of the bulb is 2.5% and it is a point source.
Solution:
The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3m, the surface area of the surrounding sphere is A = 4 πr² = 4π (3)² = 113m²
The intensity at this distance is
I = \(\frac{\text { Power }}{\text { Area }}=\frac{100 \mathrm{~W} \times 2.5 \%}{113 \mathrm{~m}^2}\) = 0.022 W.m²
Half of this intensity is provided by the electric field and half by the magnetic field.
\(\frac{1}{2} I=\frac{1}{2}\left(\varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2 \mathrm{C}\right)\)
= \(\frac{1}{2}\) (0.022 W/m²)
E_rms = \(\sqrt{\frac{0.022}{\left(8.85 \times 10^{-12}\right)\left(3 \times 10^8\right)}}\) V/m = 2.9 V/m
The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E₀ is
E₀ = \(\sqrt{2}\)E_rms = \(\sqrt{2}\) × 2.9 V/m
= 4.07 V/m
Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre.
Now, let us calculate the strength of the magnetic field. It is
B_rms = \(\frac{E_{\mathrm{rms}}}{\mathrm{C}}=\frac{2.9 \mathrm{Vm}^{-1}}{3 \times 10^8 \mathrm{~ms}^{-1}}\) = 9.6 × 10^-9 T.
Again, since tbs field in the light beam is sinusoidal, the peak magnetic field is B₀ = \(\sqrt{2}\) B_rms = 1.4 × 10^-8 T. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.

Students get through AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current

Very Short Answer Questions

Question 1.
A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the ‘ secondary if the primary has 10 turns. [T.S. Mar. 16]
Answer:
\(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)
V_p = 200V, V_s = 2000V, N_p = 10
N_s = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}\) × N_p = \(\frac{2000}{200}\) × 10
N_s = 100.

Question 2.
What type of transformer is used in a 6V bed lamp ? [A.P. Mar. 17]
Answer:
Step down transformer is used in 6V bed lamp.

Question 3.
What is the phenomenon involved in the working of a transformer ? [Mar. 16(A.P.) Mar. 14]
Answer:
Transformer works on the principle of mutual induction.

Question 4.
What is transformer ratio ?
Answer:
The ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio.
Transformer ratio = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)

Question 5.
Write the expression for the reactance of i) an inductor and (ii) a capacitor.
Answer:

1. Inductive reactance (X_L) = ωL
2. Capacitive reactance (X_C) = \(\frac{1}{\omega C}\)

Question 6.
What is the phase difference between A.C emf and current in the following: Pure resistor, pure inductor and pure capacitor. [T.S. Mar. 15]
Answer:

1. In pure resistor A.C. e.m.f and current are in phase with each other.
2. In pure inductor, current lags behind the e.m.f. by an angle of \(\frac{\pi}{2}\) (or) 90°.
3. In pure capacitor, current leads the e.m.f by an angle \(\frac{\pi}{2}\) (or) 90°.

Question 7.
Define power factor. On which factors does power factor depend ?
Answer:
The ratio of true power and apparent power (virtual power) in an a.c circuit is called as power factor of the circuit.
Power factor (cosΦ) = \(\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{rms}}}\) [∵ P_rms = V_rms I_rms]
Power factor depends on r.m.s voltage, r.m.s current and average power (P).

Question 8.
What is meant by wattless component of current ?
Answer:
Average power (P) = V_rms(I_rms sinΦ) cos\(\frac{\pi}{2}\)
The average power consumed in the circuit due to (I_rms sinΦ) component of current is zero. This component of current is known as wattless current. (I_rms sinΦ) is the wattless component of current.

Question 9.
When does a LCR series circuit have minimum impedance ?
Answer:
In LCR series circuit, Impendence (Z) = \(\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}\)
At a particular frequency, ωL = \(\frac{1}{\omega C}\)
The impedance is minimum (Z = R)
This frequency is called resonant frequency.

Question 10.
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity ?
Answer:
In LCR series circuit power factor (cosΦ) = 1
Phase difference between voltage and current is zero. (Φ = 0)

Short Answer Questions

Question 1.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Answer:
Transformer is a device to convert a low alternating current of high voltage into high alternating current of low voltage and vice versa.
Principle : It works on the principle of mutual induction between two coils.
Working : When an alternating emf is applied across the primary coil, the input voltage changes with time. Hence the magnetic flux through the primary also changes with time.

This changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.

Theory: Let N₁ and N₂ be the number of turns in the primary and secondary. Let V_P and V_S be the emf s across the primary and secondary.
\(\frac{V_S}{V_p}=\frac{\text { Output emf }}{\text{Input emf}}=\frac{-N_2 \frac{d \phi}{d t}}{-N_1 \frac{d \phi}{d t}}=\frac{N_2}{N_1}\)
∴ \(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_2}{\mathrm{~N}_1}\) = Transformer ratio
Efficiency of transformer :
It is the ratio of output power to the input power.
η = \(\frac{\text { Outputpower }}{\text { Input power }}\) × 100

Problems

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υ_m = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
X_L = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance are given by i = \(\sqrt{2}\) sin (100t – π/4)A and υ = 40 sin (100t) V. Calculate the resistance ?
Solution:
i = \(\sqrt{2}\) sin (100t – π/4)A (∵i = i₀sin(ωt – Φ))
υ = 40 sin(100t)V (∵ V = V₀sin(ωt ))
i₀ = \(\sqrt{2}\) , V₀ = 40, ω = 100, Φ = π/4
R = \(\frac{\mathrm{V}_0}{\mathrm{i}_0}\) cosΦ = \(\frac{40}{\sqrt{2}}\) cos\(\frac{\pi}{4}\), R = \(\frac{40}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\), R = 20 Ω

Question 4.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the voltage supplied by the alternator.
Solution:
V_C = 20V, V_R = 35V, V_L = 20V
V = \(\sqrt{V_R^2+\left(V_L^2-V_C^2\right)}\) ; V = \(\sqrt{(35)^2+\left(20^2-20^2\right)}\) ; V = \(\sqrt{35^2}\); V = 35 Volt.

Question 5.
What is step up transformer ? How it differs from step down transformer ?
Solution:
The ratio of number of turns in the secondary coil to the number of turns in the primary coil is called transformer ratio.
T = \(\frac{N_S}{N_p}=\frac{\text {No. of turns in the secondary }}{\text {No. of turns in the primary }}\)
If N_S > N_P, then the transformer is called step up transformer.
If N_S < N_P, then the transformer is called step down transformer.

Textual Examples

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V.
The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υ_m = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
X_L = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ?
Solution:
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

Question 4.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ?
Solution:
The capacitive reactance is
X_c = \(\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}\)
= 212 Ω
The rms current is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}=\frac{220 \mathrm{~V}}{212 \Omega}\)
= 1.04 A
The peak current is
i_m = \(\sqrt{2}\)I = (1.41)(1.04A) = 1.47A
This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 5.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.

The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons.
Solution:
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

Question 6.
A resistor of 200Ω and a capacitor of 15.0 μF are connected in series to a 220V, 50 Hz ac source,
(a) Calculate the current in the circuit;
(b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.
Solution:
Given
R = 200Ω. C = 15.0 μF = 15.0 × 10^-6F
V = 220V, v = 50Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
Z = \(\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{C}}^2}=\sqrt{\mathrm{R}^2+(2 \pi v C)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+\left(2 \times 3.14 \times 50 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+(212 \Omega)^2}\) = 291.5Ω
Therefore, the current in the circuit is
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{220 \mathrm{~V}}{291.5 \Omega}\) = 0.755A

(b) Since the current is the same throughout the circuit, we have
V_R = IR = (0.755 A) (200Ω) = 151V
V_C = IX_C = (0.755A) (212.3Ω) = 160.3V
The algebraic sum of the two voltages, V_R and V_C is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox ? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:
V_R+C = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 7.
a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
Solution:
a) We know that P = IV cosΦ where cosΦ is the power factor. To supply a given power at a given voltage, if cosΦ is small, we have to increase current accordingly. But this will lead to large power loss (I^R) in transmission.

b) Suppose in a circuit, current I lags the voltage by an angle Φ.
Then power factor cosΦ = R/Z
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram in the figure how this can be achieved. Let us resolve I into two components, I_P

along the applied voltage V and I_q perpendicular to the applied voltage. I_q is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I’_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I’_q cancel each other and P is effectively I_P V.

Question 8.
A sinusoidal voltage of peak value 283 V , and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω. L = 25.48 mH. and C = 796μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Solution:
a) To find the impedance of the circuit, we first calculate X_L and X_C.
X_L = 2πvL
= 2 × 3.14 × 50 × 25.48 × 10^-3Ω = 8Ω
X_C = \(\frac{1}{2 \pi v \mathrm{C}}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4Ω
Therefore,
z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=\sqrt{3^2+(8-4)^2}\)
= 5Ω

b) Phase difference, Φ = tan^-1\(\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
= tan^-1\(\left(\frac{4-8}{3}\right)\) = -53.1°

c) The power dissipated in the circuit is
P = I²R

Therefore, P = (40A)² × 3Ω = 4800W
= 4.8 kW

d) Power factor = cos Φ = cos 53.1° = 0.6.

Question 9.
Suppose the frequency of the source in the previous example can be varied,
(a) What is the frequency of the source at which resonance occurs ?
(b) Calculate the impedance, the current, and the power dissipated at the resonant condition.
Solution:
(a) The frequency at which the resonance occurs is
ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}}\)
= 222.1 rad/s
v_r = \(\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14}\) Hz = 35.4Hz

b) The impedance Z at resonant condition is equal to the resistance
Z = R = 3Ω
The rms current at resonance is ,
as V = \(\frac{v_{\mathrm{m}}}{\sqrt{2}}\)
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\mathrm{R}}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}\) = 66.7 A
The power dissipated at resonance is
P = I² × R = (66.7)² × 3 = 13.35 kW
You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 8.

Question 10.
At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ?
Solution:
The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in a significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

Question 11.
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Solution:
Let q₀ be the initial charge tin a capacitor. Let the charged capacitor be connected to an inductor of inductance L. this LC circuit will sustain an oscillation with frequency
\(\omega\left(2 \pi v=\frac{1}{\sqrt{\mathrm{LC}}}\right)\)
At an instant t, charge q on the capacitor and the current i are given by :
q(t) = q₀ cos ωt
i(t) = -q₀ co sin ωt
Energy stored in the capacitor at time ‘t’ is
U_E = \(\frac{1}{2}\) C V² = \(\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) cos² (ωt)
Energy stored in the inductor at time ‘t’ is
U_M = \(\frac{1}{2}\) L i²
= \(\frac{1}{2}\) L q₀² ω²sin² (ωt)
= \(\frac{Q_0^2}{2 C} \sin ^2(\omega t)\) [∵  ω = 1/ \(\sqrt{L C}\)]
Sum of energies
U_E + U_M = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) [cos² ωt + sin²ωt) = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\)
This sum is constant in time as q₀ and C, both are time-independent.

Students get through AP Inter 2nd Year Physics Important Questions 9th Lesson Electromagnetic Induction which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 9th Lesson Electromagnetic Induction

Very Short Answer Questions

Question 1.
What did the experiments of Faraday and Henry show?
Answer:
The discovery and understanding of electromagnetic induction are based on a long series of experiments carried out by Faraday and Henry.

Question 2.
Define magnetic flux.
Answer:
Magnetic flux is defined as the number of magnetic lines of force crossing through the surface.
Φ_B – \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}\) = BA cos θ
C.G.S unit → Maxwell
S.I. unit → Weber (wb)
Magnetic flux is a scalar.

Question 3.
State Faraday’s law of electromagnetic induction.
Answer:
“Magnitude of induced e.m.f is directly proportional to the rate of change of magnetic flux”
ε ∝ \(-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

Question 4.
State Lenz’s law.
Answer:
The direction of induced e.m.f (or) current is such that it opposes the cause which produces it. Lenz’s law is in accordance with law of conservation of energy.

Question 5.
What happens to the mechanical energy (of motion) when a conductor is moved in a uniform magnetic field ?
Answer:
Motional e.m.f is produced to the motion of the conductor in a magnetic field.
Motion e.m.f (ε) B/υ.

Question 6.
What are Eddy currents ? [A.P. Mar. 15]
Answer:
Eddy currents (or) Focault currents : The induced circulating currents produced in a conductor itself due to change in magnetic flux linked with the conductor are called Eddy currents.
Due to Eddy currents, the energy is dissipated in the form of heat energy.

Question 7.
Define ‘inductance’.
Answer:
Inductance is a coefficient of electromagnetic induction and is an intrinsic property of a material just like capacitance.
Inductance is an important scalar quantity which depends upon the geometry (i.e, dimensions) of a coil.

Question 8.
What do you understand by ‘self inductance’ ?
Answer:
Self inductance of a coil is defined as the induced e.m.f produced in the coil through which the rate of change of current is unity.
ε = -L \(\frac{\mathrm{dI}}{\mathrm{dt}}\); ε = -L, If \(\frac{\mathrm{dI}}{\mathrm{dt}}\) = 1 A/s.

Short Answer Questions

Question 1.
Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion.
Answer:
Consider a conductor PQ of length l moving freely in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\) with uniform veiority υ on a rectangular conductor ABCD. Let any arbitrary charge q in the conductor also move in the field with same velocity.

Magnitude of Lorentz force on this charge
(F) = Bqυ ……………….. (1)
Workdone in moving the charge from P to Q is given by ‘
W = Force × displacement
W = Bqυ × l ………………. (2) (∵ Direction of force on the charge as per Flemings left hand rule)
Electromagnetic force (ε) = \(\frac{W}{Q}\)
ε = \(\frac{\mathrm{Bqυl}}{\mathrm{q}}\) ⇒ ε = Blυ ……………. (3)

Question 2.
Describe the ways in which Eddy currents are used to advantage. [A.P. Mar. 17, 16; A.P. & T.S. Mar. 15]
Answer:
Eddy currents are used to advantage in
i) Magnetic braking in trains : A strong magnetic field is applied across the metallic drum rotating with the axle of the electric train. Thus large eddy currents are produced in the metallic drum. These currents oppose the motion of the drum and hence the axle of the train which ultimately makes the train come to rest.

ii) Induction Motor: Eddy currents are used to rotate the short circuited motor of an induction motor. Ceiling fans are also induction motors which run on single phase alternating current.

iii) Electromagnetic damping : Certain galvanometers have a fixed core made of non magnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.

iv) Induction furnace : Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil. The eddy currents generated in the metals produce high temperatures sufficient to melt it.

v) Analogue energy meters : Concept of eddy currents is used in energy meters to record the consumption of electricity. Aluminium disc used in these meters get induced due to varying magnetic field. It rotates due to eddy currents produced in it.

Question 3.
Obtain an expression for the mutual inductance of two long co-axial solenoids.
Answer:
Consider two solenoids as shown in figure. The length of primary coil be l and area of cross section A. Let N₁ and N₂ are the total number of turns in the primary and secondary solenoids. Let n₁ and n₂ be the number of turns per unit length
(n₁ = \(\frac{\mathrm{N}_1}{l}\) and n₂ = \(\frac{\mathrm{N}_2}{l}\)). Current in the primary coil is i.

∴ Magnetic field inside the primary (B) = μ₀n₁ I = μ₀ \(\frac{\mathrm{N}_1}{l}\) I ……….. (1)
Magnetic flux through each turn of primary
Φ_B = \(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}\) = μ₀ \(\frac{\mathrm{N}_1}{l}\) I × A ……………. (2)
The same magnetic flux is linked with the secondary coil.
∴ Total magnetic flux linked with secondary = μ₀\(\frac{\mathrm{N}_1 \mathrm{i}}{l}\) × A × N₂ ……………. (3)
If M be mutual inductance of the two coils, the total flux linked with the secondary is Mi.
∴ Mi = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{iA}}{l}\) ……………… (4)
M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{l}\) …………….. (5) (∵ A = πr²)
(or) M = \(\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2\left(\pi \mathrm{r}^2\right)}{l}\) ………….. (6) (∵μ_r = \(\frac{\mu}{\mu_0}\))

Problems

Question 1.
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? (Note that 1 G = 10^-4 T.)
Solution:
Induced emf = (1/2) ω B R²
= (1/2) × 4π × 0.4 × 10^-4 × (0.5)²
= 6.28 × 10^-5 V

Question 2.
Number of turns in a coil are 100. When a current of 5A is flowing through the coil, the magnetic flux is 10^-6Wb. Find the self induction. [Board Model Paper]
Solution:
Self inductance, L = \(\frac{n \phi}{\mathrm{i}}\)
number of turns, n = 100; magnetic flux, Φ = 10^-6Wb; Current, i = 5A
L = \(\frac{100 \times 10^{-6}}{5}\) = 20 × 10^-6 = 20 µH
∴ Self inductance, L = 20 µH

Question 3.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit. [Mar. 16 (T.S.) Mar. 14]
Solution:
Change in current, dI = 5 – 0 = 5A,
Time taken in current change dt = 0.1 s
Induced average emf e_av = 200 V
Induced emf in the circuit e = L . \(\frac{\mathrm{dI}}{\mathrm{dt}}\) ⇒ 200 = L\(\left(\frac{5}{0.1}\right)\) or L = \(\frac{200}{50}\) = 4 H.

Question 4.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ? [T.S. Mar. 17]
Solution:
Given, mutual inductance of coil M = 1.5 H
Current change in coil dI = 20 – 0 = 20 A
Time taken in change dt = 0.5s, Induced emf in the coil e M = \(M \frac{d I}{d t}=\frac{d \phi}{d t}\)
dΦ = M.dI = 1.5 × 20, dΦ = 30 Wb,
Thus the change of flux linkage is 30 Wb.

Question 5.
A jet plane is travelling towards west at a speed of 1800 km/K What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°.
Solution:
Speed of jet plane V = 1800 km/h = 1800 × \(\frac{5}{18}\) = 500 m/s
l = Distance between the ends of wings = 25 m
The magnitude of magnetic field B = 5 × 10^-4 T
Angle of dip γ = 30°.
Use the formula of motional emf
e = B_vVl, e = B sin γ Vl (B_v = B sin γ),
e = 5 × 10^-4 × sin 30° × 500 × 25, e = 3.1V
Thus, the voltage difference developed between the ends is 3.1 V.

Textual Examples

Question 1.
(a) What would you do to obtain a large deflection of the galvanometer ?
(b) How would you demonstrate the presence of an induced current in the absence of a gal-vanometer ?
Solution:
a) To obtain a large deflection, one or more of the following steps can be taken :

1. Use a rod made of soft iron inside the coil C₂.
2. Connect the coil to a powerful battery, and
3. Move the arrangement rapidly towards the test coil C₁.

b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.

Question 2.
A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time interval.
Solution:
The angle θ made by the area vector of the loop with the magnetic field is 45°.
From eq. Φ_B = B.A. = BA cosθ the initial magnetic flux is Φ = BA cosθ
= \(\frac{0.1 \times 10^{-2}}{\sqrt{2}}\) Wb
Final flux, Φ_min = 0
The change in flux is brought about in 0.70 s. From Eq. ε = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\), the magnitude of the induced emf is given by
ε = \(\frac{\left|\Delta \phi_{\mathrm{B}}\right|}{\Delta \mathrm{t}}=\frac{|(\phi-0)|}{\Delta \mathrm{t}}=\frac{10^{-3}}{\sqrt{2} \times 0.7}\) = 1.0 mV
And the magnitude of the current is
I = \(\frac{\varepsilon}{\mathrm{R}}=\frac{10^{-3} \cdot \mathrm{V}}{0.5 \Omega}\) = 2 mA.
Note that the earth’s magnetic field also produces a flux through the loop.

Question 3.
A circular coil of radius 10 cm, 1500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. .Horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T.
Solution:
Initial flux through the coil,
Φ_B(initial) = BA C0S θ
= 3.0 × 10^-5 × (π × 10^-2) × cos 0°
= 3π × 10^-7 Wb.
Final flux after the rotation,
Φ_B(final) = 3.0 × 10^-5 × (π × 10^-2) × cos 180°
= -3π × 10^-7 Wb.
Therefore, estimated value of the induced emf is,
ε = N\(\frac{\Delta \phi}{\Delta t}\) = 500 × (6π × 10^-7)/0.25
= 3.8 × 10^-3 V
I = ε/R = 1.9. × 10^-3 A.
Note that the magnitudes of ε and I are the estimated values.

Question 4.
The following figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.

Solution:

1. The magnetic flux through the rectangular loop abed increases, due to the motion of the loop into the region of magnetic field. The induced current must flow along the path bcdab so that it opposes the increasing flux.
2. Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacd, so as to oppose the change in flux.
3. As the magnetic flux decreases due to motion of the irregular shaped loop abed out of the region of magnetic field, the induced current flows along edabe, so as to oppose change in flux.

Question 5.
a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets ?
b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) When it is partially outside the plates of the capacitor ? The electric field is normal to the plane of the loop.
c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region as in the figure, to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region ? The field is normal to the loops.

d) Predict the polarity of the capacitor in the situation described by the following figure.

Solution:
a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.

b) No current is induced in either case. Current can not be induced by changing the electric flux.

c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.

d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Question 6.
A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring as in figure. A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring ?

Solution:
Method I : As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.
Using equation (ε = – Bl \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = BlV), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by
dε = Bυ dr. Hence,
ε = \(\int \mathrm{d} \varepsilon=\int_0^{\mathrm{R}} \mathrm{B} v \mathrm{dr}=\int_0^{\mathrm{R}} \mathrm{B} \omega \mathrm{rdr}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)
Note that we have used υ = ωr. This gives
ε = \(\frac{1}{2}\) × 1.0 × 2π × 50 × (1²) = 157 V.

Method II: To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by
πR² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\), R²θ
where R is the radius of the circle. Hence, the induced emf is
e = B × \(\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{1}{2} \mathrm{R}^2 \theta\right]=\frac{1}{2} \mathrm{BR}^2 \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)
[Note: \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = ω = 2πv]
This expression is identical to the – expression obtained by Method I and we get the same value of ε.

Question 7.
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_E at a place. If H_E = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? Note that 1 G = 10^-4 T.
Solution:
Induced emf = (1/2) ω B R²
= (1/2) × 4π × 0.4 × 10^-4 × (0.5)²
= 6.28 × 10^-5 V

Question 8.
Refer to fig. The arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed o. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

Solution:
Let us first consider the forward motion from x = 0 to x = 2b. The flux Φ_B linked with the circuit SPQR is
Φ_B = Blx 0 ≤ x < b
= Blυ b ≤ x < 2b
The induced emf is,
ε = – \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
= -Blυ 0 ≤ x < b
= 0 b ≤ x < 2b.
When the induced emf is non-zero, the current I is (in magnitude)
I = \(\frac{\mathrm{B} l v}{\mathrm{r}}\)

The force required to keep the arm PQ in constant motion is I lB. Its direction is to the left. In magnitude
F = \(\frac{\mathrm{B}^2 l^2 v}{\mathrm{r}}\) 0 ≤ x < b
= 0 b ≤ x < 2b
The Joule heating loss is
P_j = I²r
= \(\frac{\mathrm{B}^2 l^2 v^2}{\mathrm{r}}\)   0 ≤ x < b
= 0   b ≤ x < 2b
One obtains similar expressions for the inward motion from x = 2b to x = 0.

Question 9.
Two concentric circular coils, one of small radius r₁ and the other of large radius r₂, such that r₁ << r₂, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.
Solution:
Let a current I₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = μ₀I₂/2r₂. Since the other co-axially placed coil has a very small radius. B₂ may be considered constant over its cross-sectional area. Hence,
Φ₂ = \(\pi \mathrm{r}_1^2 \mathrm{~B}_2\)
= \(\frac{\mu_0 \pi \mathrm{r}_1^2}{2 \mathrm{r}_2} \mathrm{I}_2\)
= M₁₂I₂
Thus,
M₁₂ = \(\frac{\mu_0 \pi r_1^2}{2 r_2}\)
From Equation M₁₂ = M₂₁ = M
M₁₂ = M₂₁ = \(\frac{\mu_0 \pi r_1^2}{2 r_2}\)
Note that we calculated M₁₂ from an approximate value of Φ₁ assuming the magnetic field B₂ to be uniform over the area \(\pi r_1^2\). However, we can accept this value because r₁ << r₂.

Question 10.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid.
(b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor ?
Solution:
a) From Equation ε = – L \(\frac{\mathrm{dI}}{\mathrm{dT}}\),
the magnetic energy is
U_B = \(\frac{1}{2}\) LI²
= \(\frac{1}{2} \mathrm{~L}\left(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\right)^2\)
(since B = μ₀nI, for a solenoid)
= \(\frac{1}{2}\) (μ₀n²Al) \(\left(\frac{\mathrm{B}}{\mu_0 \mathrm{n}}\right)^2\)
[from Equation L = μ₀n²Al]
= \(\frac{1}{2 \mu_0}\) B²Al

b) The magnetic energy per unit volume is,
u_B = \(\frac{\mathrm{U}_{\mathrm{B}}}{\mathrm{V}}\)
[where Vis volume that contains flux]
= \(\frac{\mathrm{U}_{\mathrm{B}}}{\mathrm{Al}}=\frac{\mathrm{B}}{2 \mu_0}\) ……………….. (1)
We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor.
u_E = \(\frac{1}{2}\) ε₀E² ………………. (2)
In both the cases energy is proportional to the square of the field strength.

Question 11.
Kamla peddles a stationary bicycle, the pedals of the bicycle are attached to a 100 turn coil of area 0.10 m². The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil ?
Solution:
Here f = 0.5 Hz; N = 100, A = 0.1 m² and B = 0.01 T.
Employing Equation ε = NBA ω sin ωt
ε₀ = NBA (2πv)
= 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5
= 0.314 V
The maximum voltage is 0.314 V