CBSE Class 12 Physics Chapter 6: Electromagnetic Induction - NCERT Solutions

The problem involves a rectangular loop in a changing magnetic field, leading to induced emf and power dissipation.

Given:

• Sides of the rectangular loop are 8 cm and 2 cm.
• Initial magnetic field, \( B_{initial} = 0.3 \) T.
• Rate of decrease of magnetic field, \( \frac{dB}{dt} = -0.02 \) T/s (negative sign indicates decrease).
• Resistance of the loop, \( R = 1.6 \, \Omega \).

First, calculate the area of the rectangular loop:

\( A = \text{length} \times \text{width} = 8 \, \text{cm} \times 2 \, \text{cm} = 16 \, \text{cm}^2 \)

Convert the area to square meters:

\( A = 16 \times 10^{-4} \, \text{m}^2 \)

According to Faraday's Law of Induction, the magnitude of the induced emf (\( e \)) in the loop is given by the rate of change of magnetic flux (\( \phi \)) through it:

\( e = \left| \frac{d\phi}{dt} \right| \)

The magnetic flux is given by \( \phi = AB \), where \( A \) is the area and \( B \) is the magnetic field strength. Since the area \( A \) is constant and the magnetic field \( B \) is changing with time, we have:

\( e = \left| \frac{d(AB)}{dt} \right| = A \left| \frac{dB}{dt} \right| \)

Substitute the given values:

\( e = (16 \times 10^{-4} \, \text{m}^2) \times | -0.02 \, \text{T/s} | = 16 \times 10^{-4} \times 0.02 \, \text{V} \)

\( e = 0.32 \times 10^{-4} \, \text{V} \)

Now, calculate the induced current (\( i \)) in the loop using Ohm's Law (\( i = \frac{e}{R} \)):

\( i = \frac{0.32 \times 10^{-4} \, \text{V}}{1.6 \, \Omega} = 0.2 \times 10^{-4} \, \text{A} = 2 \times 10^{-5} \, \text{A} \)

The power (\( P \)) dissipated by the loop as heat is given by \( P = i^2 R \):

\( P = (2 \times 10^{-5} \, \text{A})^2 \times 1.6 \, \Omega = (4 \times 10^{-10} \, \text{A}^2) \times 1.6 \, \Omega \)

\( P = 6.4 \times 10^{-10} \, \text{W} \)

Source of Power: The power dissipated as heat originates from the work done by an external agent that is responsible for gradually reducing the magnetic field. This work is converted into electrical energy, which then dissipates as heat due to the resistance of the loop, in accordance with the conservation of energy.

This problem involves calculating the induced current in a loop moving through a magnetic field that is non-uniform in space and also changing with time.

Given:

• Side of the square loop, \( s = 12 \, \text{cm} = 0.12 \, \text{m} \).
• Velocity of the loop, \( v = 8 \, \text{cm/s} = 0.08 \, \text{m/s} \) in the positive x-direction.
• Magnetic field gradient along the negative x-direction, \( \frac{dB}{dx} = 10^{-3} \, \text{T/cm} \). Converting this to T/m: \( \frac{dB}{dx} = 10^{-3} \, \text{T/(10}^{-2} \, \text{m)} = 10^{-1} \, \text{T/m} \). This means the field increases by \( 0.1 \) T for every meter moved in the negative x-direction.
• Rate of decrease of the magnetic field with time, \( \frac{dB}{dt} = -10^{-3} \, \text{T/s} \) (negative sign indicates decrease).
• Resistance of the loop, \( R = 4.50 \, \text{m}\Omega = 4.50 \times 10^{-3} \, \Omega \).

The area of the square loop is \( A = s^2 = (0.12 \, \text{m})^2 = 0.0144 \, \text{m}^2 \).

The total induced emf (\( e \)) in the loop is due to two effects: the change in magnetic field with time and the change in magnetic field with position as the loop moves.

1. Emf due to time variation of the magnetic field (\( e_t \)):

This is given by Faraday's Law: \( e_t = A \left| \frac{dB}{dt} \right| \).

\( e_t = (0.0144 \, \text{m}^2) \times |-10^{-3} \, \text{T/s}| = 0.0144 \times 10^{-3} \, \text{V} \).

2. Emf due to motion in a non-uniform magnetic field (\( e_m \)):

As the loop moves in the positive x-direction with velocity \( v \), the magnetic field strength changes across its width. The change in magnetic field experienced by the loop due to its motion is related to the gradient \( \frac{dB}{dx} \). The effective change in B across the loop's dimension in the x-direction (which is the side length \( s \)) is \( s \frac{dB}{dx} \). The motional emf is then \( e_m = v \times (\text{change in B across the loop}) \). Since the field increases in the negative x-direction, as the loop moves in the positive x-direction, the field it enters is weaker than the field it leaves. The emf generated is:

\( e_m = v \times s \times \frac{dB}{dx} \)

\( e_m = (0.08 \, \text{m/s}) \times (0.12 \, \text{m}) \times (0.1 \, \text{T/m}) \)

\( e_m = 0.00096 \, \text{V} = 0.96 \times 10^{-3} \, \text{V} \).

The total induced emf is the sum of these two components. However, we need to consider the direction. The magnetic field is in the +z direction. The loop moves in the +x direction. The field gradient is such that B increases as x decreases. As the loop moves in +x, it encounters regions of weaker B. The change in flux is due to both time variation and spatial variation. The total emf is given by \( e = A \frac{dB}{dt} + v L B_x \), where \( L \) is the length perpendicular to motion and field, and \( B_x \) is the gradient term. A more direct way for motional emf in a gradient is \( e_m = \int (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \). For a rectangular loop moving in the x-direction through a B field in the z-direction with a gradient in x, the emf across the sides parallel to the y-axis will differ. Let the field at position x be \( B(x, t) \). The flux is \( \phi = A B(x, t) \). The total emf is \( e = -\frac{d\phi}{dt} = - \frac{d}{dt} (A B(x, t)) = -A \frac{dB}{dt} \). However, the velocity also changes the flux. The change in flux due to motion is \( \Delta \phi = \int (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{A} \). For a loop moving in +x direction, the emf induced is \( e = v \times (\text{width}) \times (\text{gradient of B along x}) \). The width is \( s \). So \( e_m = v \times s \times \frac{dB}{dx} \). The total emf is the sum of the emf due to time change and motional emf in the gradient field.

Total emf \( e = e_t + e_m \). We need to be careful about signs. The magnetic field is in +z. Velocity is in +x. Gradient \( \frac{dB}{dx} \) is positive for increase in B as x decreases. As loop moves in +x, it enters region of weaker B. The induced current will try to oppose this decrease. The rate of change of flux is \( \frac{d\phi}{dt} = A \frac{dB}{dt} + (\mathbf{v} \times \mathbf{B}) \cdot

abla A \). A simpler approach is to consider the emf induced in the sides parallel to the y-axis. Let the field at the center of the loop be \( B_0 \). The field varies as \( B(x) = B_0 + x \frac{dB}{dx} \). The loop extends from \( x \) to \( x+s \). The flux is \( \phi = s \times s \times B_{avg} \). The emf is \( e = -\frac{d\phi}{dt} \). The total emf is the sum of the emf due to time variation and motional emf. The motional emf is \( e_m = \int (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \). For the sides parallel to y-axis, \( d\mathbf{l} = dy \hat{j} \). \( \mathbf{v} = v \hat{i} \). \( \mathbf{B} = B \hat{k} \). \( \mathbf{v} \times \mathbf{B} = v \hat{i} \times B \hat{k} = -vB \hat{j} \). So \( (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} = -vB dy \). Integrating from \( y=0 \) to \( y=s \), we get \( -vBs \). This is for one side. For the other side at \( x+s \), the field is \( B + s \frac{dB}{dx} \). The emf is \( -v(B+s\frac{dB}{dx})s \). The net emf is \( -vBs - (-v(B+s\frac{dB}{dx})s) = vs \frac{dB}{dx} \). This is the motional emf. The total emf is \( e = A \frac{dB}{dt} + vs \frac{dB}{dx} \). Using the given values:

\( e = (0.0144 \, \text{m}^2) \times (-10^{-3} \, \text{T/s}) + (0.08 \, \text{m/s}) \times (0.12 \, \text{m}) \times (0.1 \, \text{T/m}) \)

\( e = -0.0144 \times 10^{-3} \, \text{V} + 0.00096 \, \text{V} \)

\( e = -0.0144 \times 10^{-3} \, \text{V} + 0.96 \times 10^{-3} \, \text{V} \)

\( e = (0.96 - 0.0144) \times 10^{-3} \, \text{V} = 0.9456 \times 10^{-3} \, \text{V} \).

The magnitude of the induced emf is \( |e| = 0.9456 \times 10^{-3} \, \text{V} \).

The magnitude of the induced current (\( i \)) is given by \( i = \frac{|e|}{R} \):

\( i = \frac{0.9456 \times 10^{-3} \, \text{V}}{4.50 \times 10^{-3} \, \Omega} \approx 0.2101 \, \text{A} \).

Direction of Induced Current: The net change in flux is positive (since the motional emf term is larger and positive, indicating an increase in flux in the direction that opposes the change). According to Lenz's Law, the induced current will flow in a direction that opposes this increase in flux. Since the magnetic field is in the +z direction, the induced current will create a magnetic field in the -z direction. This corresponds to a counter-clockwise current when viewed from the positive z-axis (i.e., clockwise when viewed from the negative z-axis).

Final Answer: The magnitude of the induced current is approximately 0.21 A, and its direction is clockwise when viewed from the negative z-axis (or counter-clockwise when viewed from the positive z-axis).