Chapter 6 Electromagnetic Induction Class 12 Physics

Question 6.1

Predict the direction of induced current in the situations described by the following figures (a) to (f).

Concept: Lenz’s Law

The direction of induced current is such that it opposes the change in magnetic flux that produced it.

[Image of Lenz’s Law Current Direction]

Analysis

(a) South pole enters coil: Induced current creates a South pole (Clockwise from right) to repel it. Direction: qrpq.
(b) South pole leaves first coil: Induced pole is North (attract). Direction: prq. North pole enters second coil: Induced pole is North (repel). Direction: yzx.
(c) Tapping key closed: Flux increases. Induced current creates opposing field. Direction: yzx.
(d) Rheostat setting changed (Current I decreases): Flux decreases. Induced current tries to maintain flux. Direction: zyx.
(e) Tapping key released: Flux decreases. Induced current opposes decay. Direction: xry.
(f) Current I decreases: Field lines shrink. Induced lines support main field. Current flows along the field lines of the straight wire. No induced current in the loop axis.

Question 6.2

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Formula

Motional EMF for a rotating rod: $\varepsilon = \frac{1}{2} B \omega L^2$.

Given

$L = 1.0 \text{ m}, \omega = 400 \text{ rad/s}, B = 0.5 \text{ T}$.

Calculation

$$\varepsilon = \frac{1}{2} \times 0.5 \times 400 \times (1.0)^2$$ $$\varepsilon = 0.5 \times 200 = 100 \text{ V}$$

Result: 100 V

Question 6.3

A long solenoid with 15 turns per cm has a small loop of area $2.0 \text{ cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

1. Magnetic Field of Solenoid

$n = 15 \text{ turns/cm} = 1500 \text{ turns/m}$.
$B = \mu_0 n I$.

2. Rate of Change of Flux

$\Phi = B A = (\mu_0 n I) A$.
Rate of change $\varepsilon = \frac{d\Phi}{dt} = \mu_0 n A \frac{dI}{dt}$.

$$\frac{dI}{dt} = \frac{4.0 – 2.0}{0.1} = 20 \text{ A/s}$$ $$\varepsilon = 4\pi \times 10^{-7} \times 1500 \times (2.0 \times 10^{-4}) \times 20$$ $$\varepsilon = 4\pi \times 10^{-7} \times 6 \times 10^3 \times 10^{-4} \times 10$$ $$\varepsilon = 24\pi \times 10^{-7} \approx 7.5 \times 10^{-6} \text{ V}$$

Result: $7.5 \times 10^{-6} \text{ V}$

Question 6.4

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm/s in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Concept

EMF $\varepsilon = B L v$, where $L$ is the length of the side perpendicular to velocity (cutting the field lines).

(a) Velocity normal to Longer Side (8 cm)

Here, the effective length cutting the field is the longer side $L = 8 \text{ cm} = 0.08 \text{ m}$.
$v = 0.01 \text{ m/s}, B = 0.3 \text{ T}$.

$$\varepsilon = 0.3 \times 0.08 \times 0.01 = 2.4 \times 10^{-4} \text{ V}$$ $$\text{Time } t = \frac{\text{Distance (Width)}}{\text{Speed}} = \frac{0.02}{0.01} = 2 \text{ s}$$

(b) Velocity normal to Shorter Side (2 cm)

Here, effective length $L = 2 \text{ cm} = 0.02 \text{ m}$.

$$\varepsilon = 0.3 \times 0.02 \times 0.01 = 0.6 \times 10^{-4} \text{ V}$$ $$\text{Time } t = \frac{\text{Distance (Length)}}{\text{Speed}} = \frac{0.08}{0.01} = 8 \text{ s}$$

Question 6.5

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod passing through its one end… Calculate the emf developed.

(Note: This is a repetition of Q6.2 in the NCERT text).

Calculation

$$\varepsilon = \frac{1}{2} B \omega L^2 = \frac{1}{2} \times 0.5 \times 400 \times 1^2 = 100 \text{ V}$$

Result: 100 V

Question 6.6

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad/s in a uniform horizontal magnetic field of magnitude $3.0 \times 10^{-2} \text{ T}$. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance $10 \Omega$, calculate the maximum current and average power loss.

1. Max EMF

$\varepsilon_{max} = N A B \omega$. Area $A = \pi r^2 = \pi (0.08)^2 \approx 0.02 \text{ m}^2$.

$$\varepsilon_{max} = 20 \times 0.02 \times 3.0 \times 10^{-2} \times 50 = 0.60 \text{ V}$$

2. Average EMF

Over a full cycle, average induced EMF is Zero.

3. Max Current

$$I_{max} = \frac{\varepsilon_{max}}{R} = \frac{0.60}{10} = 0.06 \text{ A}$$

4. Average Power

$$P_{avg} = \frac{1}{2} \varepsilon_{max} I_{max} = \frac{1}{2} \times 0.60 \times 0.06 = 0.018 \text{ W}$$

Source: The external mechanical work (rotor) provides this power.

Question 6.7

A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field, $0.30 \times 10^{-4} \text{ Wb m}^{-2}$.
(a) Instantaneous emf?
(b) Direction of emf?
(c) Which end is at higher potential?

(a) Instantaneous EMF

$$\varepsilon = B_H L v = 0.30 \times 10^{-4} \times 10 \times 5.0 = 1.5 \times 10^{-3} \text{ V} = 1.5 \text{ mV}$$

(b) Direction

Using Fleming’s Right-Hand Rule: Motion is Down, Field is North (Horizontal). Induced current flows from West to East.

(c) Polarity

In a source, current flows from low to high potential. Since current flows West to East, the East end (output) is generally considered positive in external circuit logic, but inside the wire (source), electrons move East to West, making the West end accumulate positive charge relative to East?
Wait, standard convention: If current flows West to East, West is low, East is high potential (like a battery terminal). Result: West end is at higher potential. (Using $\vec{F} = q(\vec{v} \times \vec{B})$, positive charges are forced towards the West end).

Question 6.8

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Formula

$\varepsilon = -L \frac{dI}{dt} \implies L = \frac{|\varepsilon|}{|dI/dt|}$.

Calculation

$$\frac{dI}{dt} = \frac{0 – 5.0}{0.1} = -50 \text{ A/s}$$ $$L = \frac{200}{50} = 4 \text{ H}$$

Self-Inductance = 4 Henry

NCERT Solutions