Motion of Square Loop Entering Magnetic Field Region

To determine which statements are correct, we analyze the motion of the conducting square loop as it enters the magnetic field region $y \ge 0$.

1. Equation of Motion

Let $y$ be the displacement of the leading edge of the loop inside the magnetic field region ($0 \le y \le L$).

As the loop enters the magnetic field $\vec{B} = B_0 \hat{k}$ with velocity $\vec{v} = v\hat{j}$, the magnetic flux through the loop is: $\Phi_B = B_0 L y$

The magnitude of the induced electromotive force (EMF) is given by Faraday's law: $\mathcal{E} = \left| \frac{d\Phi_B}{dt} \right| = B_0 L \frac{dy}{dt} = B_0 L v$

The induced current in the loop of resistance $R$ is: $I = \frac{\mathcal{E}}{R} = \frac{B_0 L v}{R}$

By Lenz's law, this current causes a magnetic force that opposes the motion. The magnetic force acts on the leading edge of length $L$: $F_{net} = - I L B_0 = - \frac{B_0^2 L^2 v}{R}$

Using Newton's second law, $M \frac{dv}{dt} = F_{net}$: $M \frac{dv}{dt} = - \frac{B_0^2 L^2}{R} v$

Given $K = \frac{B_0^2 L^2}{RM}$, the differential equation for velocity becomes: $\frac{dv}{dt} = - K v \quad \text{--- (1)}$

2. Velocity as a Function of Position and Time

From equation (1), expressing $\frac{dv}{dt} = v \frac{dv}{dy}$: $v \frac{dv}{dy} = - K v \implies \frac{dv}{dy} = - K \quad (\text{for } v > 0)$

Integrating with initial conditions $v = v_0$ at $y = 0$: $\int_{v_0}^{v} dv = - K \int_{0}^{y} dy \implies v(y) = v_0 - K y \quad \text{--- (2)}$

Integrating equation (1) with respect to time $t$: $\int_{v_0}^{v} \frac{dv}{v} = - K \int_0^t dt \implies v(t) = v_0 e^{-Kt} \quad \text{--- (3)}$

The position $y(t)$ as a function of time is: $y(t) = \int_0^t v(t') dt' = \int_0^t v_0 e^{-Kt'} dt' = \frac{v_0}{K} \left(1 - e^{-Kt}\right) \quad \text{--- (4)}$

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3. Evaluation of Options

• Option A: If the loop stops at distance $y_{\text{stop}}$, then setting $v(y_{\text{stop}}) = 0$ in equation (2): $y_{\text{stop}} = \frac{v_0}{K}$ For $v_0 = 1.5 KL$: $y_{\text{stop}} = \frac{1.5 KL}{K} = 1.5 L$ Since $1.5 L > L$, the loop reaches $y = L$ with a non-zero velocity ($v = 0.5 KL$) and completely enters the field region without stopping inside $0 \le y < L$. Thus, Option A is incorrect.

• Option B: When the entire loop is inside the region $y \ge 0$, the magnetic flux through the loop is constant ($\Phi_B = B_0 L^2$). Therefore, $\frac{d\Phi_B}{dt} = 0$, so no current is induced ($I = 0$). With no induced current, the net magnetic force on the loop is zero. Thus, Option B is correct.

• Option C: From equation (3), the velocity decays exponentially: $v(t) = v_0 e^{-Kt}$. The loop comes to rest ($v = 0$) asymptotically as $t \to \infty$, not at any finite time $t$. Thus, Option C is incorrect.

• Option D: To find the time $t$ when the complete loop enters the magnetic field region ($y = L$), we set $y(t) = L$ in equation (4): $L = \frac{v_0}{K} \left(1 - e^{-Kt}\right)$ Given $v_0 = 3 KL$: $L = \frac{3KL}{K} \left(1 - e^{-Kt}\right) \implies L = 3L \left(1 - e^{-Kt}\right)$ $\frac{1}{3} = 1 - e^{-Kt} \implies e^{-Kt} = \frac{2}{3}$ Taking the natural logarithm on both sides: $-Kt = \ln\left(\frac{2}{3}\right) \implies t = \frac{1}{K} \ln\left(\frac{3}{2}\right)$ Thus, Option D is correct.

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Conclusion

The correct options are B and D.