Question

A conducting square loop of side $ L $, mass $ M $, and resistance $ R $ is moving in the $ XY $ plane with its edges parallel to the $ X $ and $ Y $ axes. The region $ y \geq 0 $ has a uniform magnetic field, $ \vec{B} = B_0 \hat{k} $. The magnetic field is zero everywhere else. At time $ t = 0 $, the loop starts to enter the magnetic field with an initial velocity $ v_0 \hat{j} \, \text{m/s} $, as shown in the figure. Considering the quantity $ K = \frac{B_0^2 L^2}{RM} $ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:

• A. If \( v_0 = 1.5KL \), the loop will stop before it enters completely inside the region of magnetic field.
• B. When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
• C. If \( v_0 = \frac{KL}{10} \), the loop comes to rest at \( t = \left(\frac{1}{K}\right) \ln\left(\frac{5}{2}\right) \).
• D. If \( v_0 = 3KL \), the complete loop enters inside the region of magnetic field at time \( t = \left(\frac{1}{K}\right) \ln\left(\frac{3}{2}\right) \).

Hint:

When a loop enters a magnetic field, calculate induced emf and resulting force using \( F = -KMv \), and solve using exponential decay to find position and velocity over time.

Answer 1

The Correct Option is B, D

Background Physics:

• Faraday's Law: As the loop enters the magnetic field, there's a changing magnetic flux, inducing an emf (electromotive force) in the loop.
• Ohm's Law: The induced emf drives a current through the loop.
• Lenz's Law: The induced current creates a magnetic force that opposes the motion of the loop entering the field. This is the retarding force.
• Force on a current-carrying wire: \( F = I (\mathbf{L} \times \mathbf{B}) \), where \( \mathbf{L} \) is the length of the wire in the field.
• Newton's Second Law: The net force on the loop affects its acceleration and velocity.

Derivation of Equation of Motion

Let \( y \) be the distance the loop has entered the magnetic field at time \( t \).

1. Induced EMF: \( \varepsilon = B L v \), where \( v = \frac{dy}{dt} \).
2. Induced Current: \( I = \frac{\varepsilon}{R} = \frac{B L v}{R} \).
3. Magnetic Force: \( F = I L B = \frac{B L v}{R} L B = \frac{B^2 L^2 v}{R} = K M v \) (with \( K = \frac{B^2L^2}{RM} \)).
4. Equation of Motion: \( M \frac{dv}{dt} = -K M v \Rightarrow \frac{dv}{dt} = -K v \).
5. Solving for \( v(t) \): \( v(t) = v_0 e^{-Kt} \).
6. Solving for \( y(t) \): \( y(t) = \frac{v_0}{K} (1 - e^{-Kt}) \).

Statement A: If \( v_0 = 1.5KL \), the loop will stop before entering completely into the field.

We know that as \( t \to \infty \), \( y(t) \to \frac{v_0}{K} \). For \( v_0 = 1.5KL \), we get \( y(\infty) = 1.5L \), so the loop does enter fully. A is incorrect.

Statement B: When the entire loop is inside the magnetic field, the induced emf becomes zero since there's no change in flux. Hence, current and magnetic force also become zero. B is correct.

Statement C: Analyzing the given \( v_0 = \frac{KL}{10} \) at \( t = \frac{1}{K} \ln\left(\frac{5}{2}\right) \):

\[ y(t) = \left(\frac{v_0}{K}\right)\left(1 - e^{-Kt}\right) = \left(\frac{L}{10}\right)\left(1 - \frac{2}{5}\right) = \frac{L}{10} \cdot \frac{3}{5} = \frac{3L}{50} \]

This is far less than \( L \), so the loop has barely entered. C is incorrect.

Statement D: To find when \( y(t) = L \), set:

\[ L = \left(\frac{3KL}{K}\right)\left(1 - e^{-Kt}\right) = 3L(1 - e^{-Kt}) \Rightarrow \frac{1}{3} = 1 - e^{-Kt} \Rightarrow e^{-Kt} = \frac{2}{3} \]

\[ \Rightarrow t = \frac{1}{K} \ln\left(\frac{3}{2}\right) \]

Thus, D is correct.