Question 5

Let $y$ be the length of the loop that has entered the magnetic field. The induced emf is $\epsilon = B_0 L v$, and the induced current is $I = \frac{\epsilon}{R} = \frac{B_0 L v}{R}$. The magnetic force on the loop is $F = -ILB_0 = -\frac{B_0^2 L^2 v}{R} = -MKv$, where $K = \frac{B_0^2 L^2}{RM}$.

1. Equation of motion: $M\frac{dv}{dt} = -MKv \Rightarrow \frac{dv}{dt} = -Kv$. Integrating gives $v(t) = v_0 e^{-Kt}$.
2. In terms of position: $v\frac{dv}{dy} = -Kv \Rightarrow \frac{dv}{dy} = -K$. Integrating from $y=0$ to $y$ gives $v = v_0 - Ky$.

Analyze options: (A) To stop before entering completely ($y < L$), the velocity must reach zero at $y < L$. $0 = v_0 - Ky \Rightarrow y_{stop} = \frac{v_0}{K}$. For $v_0 = 1.5KL$, $y_{stop} = 1.5L > L$. Thus it enters completely. Incorrect. (B) When the loop is fully inside, the magnetic flux $\Phi = B_0 L^2$ is constant. Thus $\frac{d\Phi}{dt} = 0$, $I = 0$, and the magnetic force is zero. Correct. (C) Based on $v(t) = v_0 e^{-Kt}$, the velocity mathematically reaches zero only as $t \to \infty$. There is no finite time at which it 'comes to rest'. Incorrect. (D) For $v_0 = 3KL$, the loop enters completely when $y = L$. At this point, $v = 3KL - K(L) = 2KL$. Using $v(t) = v_0 e^{-Kt}$, we have $2KL = 3KL e^{-Kt} \Rightarrow e^{-Kt} = \frac{2}{3} \Rightarrow t = \frac{1}{K} \ln(\frac{3}{2})$. Correct.