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Q.5 A conducting square loop of side

$L$ , mass $M$ , and resistance $R$ is moving in the $X Y$ 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 $v_{0} \hat{j}$ m/s, as shown in the figure. Considering the quantity $K = \frac{B_{0}^{2} L^{2}}{R M}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct?

Options:
(A) If $v_{0} = 1.5 K L$ , the loop will stop before it enters completely inside the region of the magnetic field.
(B) When the complete loop is inside the region of the magnetic field, the net force acting on the loop is zero.
(C) If $v_{0} = \frac{K L}{10}$ , the loop comes to rest at $t = (\frac{1}{K}) \ln (\frac{5}{2})$ .
(D) If $v_{0} = 3 K L$ , the complete loop enters inside the region of the magnetic field at time $t = (\frac{1}{K}) \ln (\frac{3}{2})$ .

Answer: B, D

Solution:

1. Understanding the Problem

A conducting square loop of side $L$ , mass $M$ , and resistance $R$ moves in the $X Y$ -plane.
A magnetic field $\vec{B} = B_{0} \hat{k}$ exists only in the region $y \geq 0$ .
The loop starts entering the field at $t = 0$ with initial velocity $v_{0} \hat{j}$ .
The quantity $K = \frac{B_{0}^{2} L^{2}}{R M}$ is defined for simplification.
We need to evaluate the correctness of statements (A)-(D).

2. Key Physics Concepts

Magnetic Force: When the loop enters the field, a current is induced due to the change in magnetic flux, leading to a retarding force.
Faraday's Law: The induced e.m.f. $E = - \frac{d Φ_{B}}{d t}$ , where $Φ_{B}$ is the magnetic flux.
Lenz's Law: The induced current opposes the motion of the loop, causing deceleration.
Net Force Inside the Field: When the loop is entirely inside the field, the flux is constant, so no e.m.f. is induced, and the net force is zero.

3. Step-by-Step Analysis

Step 1: Induced e.m.f. and Current

As the loop enters the field, the area exposed to $\vec{B}$ changes with time. Let $y (t)$ be the position of the leading edge of the loop at time $t$ .

Flux Through the Loop:
$Φ_{B} = B_{0} \times Area = B_{0} L y (t)$
Induced e.m.f.:
$E = - \frac{d Φ_{B}}{d t} = - B_{0} L \frac{d y}{d t} = - B_{0} L v (t)$
Induced Current:
$I = \frac{E}{R} = - \frac{B_{0} L v (t)}{R}$

Step 2: Retarding Force

The magnetic force on the loop opposes its motion:

\vec{F} = I L \vec{B} \times length = I L B_{0} \hat{j} \times L \hat{k} = I L^{2} B_{0} (- \hat{i})

However, the net force is along $- \hat{j}$ because only the horizontal segments of the loop experience force (vertical segments cancel out). The correct force is:

F = I L B_{0} = - \frac{B_{0}^{2} L^{2}}{R} v (t)

This acts to decelerate the loop:

M \frac{d v}{d t} = - \frac{B_{0}^{2} L^{2}}{R} v (t)

\frac{d v}{d t} = - K v (t), where K = \frac{B_{0}^{2} L^{2}}{R M}

Step 3: Solve the Differential Equation

The equation of motion is:

\frac{d v}{d t} = - K v (t)

This is a first-order linear ODE with solution:

v (t) = v_{0} e^{- K t}

The position $y (t)$ is obtained by integrating $v (t)$ :

y (t) = \int_{0}^{t} v (t^{'}) d t^{'} = \frac{v_{0}}{K} (1 - e^{- K t})

Step 4: Evaluate the Statements

Statement (A):
"If $v_{0} = 1.5 K L$ , the loop will stop before it enters completely inside the region of the magnetic field."

The loop stops when $v (t) = 0$ , which occurs asymptotically as $t \to \infty$ .
The distance traveled is:
$y_{max} = \frac{v_{0}}{K} = \frac{1.5 K L}{K} = 1.5 L$
Since the loop has side length $L$ , it fully enters when $y = L$ . Here, $y_{max} = 1.5 L > L$ , so the loop does not stop before fully entering.
Conclusion: False.

Statement (B):
"When the complete loop is inside the region of the magnetic field, the net force acting on the loop is zero."

When fully inside, the flux $Φ_{B} = B_{0} L^{2}$ is constant, so $E = 0$ , $I = 0$ , and $F = 0$ .
Conclusion: True.

Statement (C):
"If $v_{0} = \frac{K L}{10}$ , the loop comes to rest at $t = (\frac{1}{K}) \ln (\frac{5}{2})$ ."

The loop never truly "comes to rest" ( $v (t) \to 0$ only as $t \to \infty$ ).
The time to reach a negligible velocity is not finite.
Conclusion: False.

Statement (D):
"If $v_{0} = 3 K L$ , the complete loop enters inside the region of the magnetic field at time $t = (\frac{1}{K}) \ln (\frac{3}{2})$ ."

The loop fully enters when $y (t) = L$ :
$L = \frac{3 K L}{K} (1 - e^{- K t}) ⟹ 1 = 3 (1 - e^{- K t})$ $e^{- K t} = \frac{2}{3} ⟹ t = \frac{1}{K} \ln (\frac{3}{2})$
Conclusion: True.

4. Final Answer

The correct statements are B and D.

\boxed{B, D}