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, $\vec{B}=B_0 k$. The magnetic field is zero everywhere else. At time $t=0$, the loop starts to enter the magnetic field with an initial velocity $v_0 \hat{\jmath} \mathrm{~m} / \mathrm{s}$, 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:

• A

  If $v_0=1.5 K L$, the loop will stop before it enters completely inside the region of magnetic field.
• Correct answer
  B

  When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
• C

  If $v_0=\frac{K L}{10}$, the loop comes to rest at $t=\left(\frac{1}{K}\right) \ln \left(\frac{5}{2}\right)$.
• Correct answer
  D

  If $v_0=3 K L$, the complete loop enters inside the region of magnetic field at time $t=\left(\frac{1}{K}\right) \ln \left(\frac{3}{2}\right)$.

Solution

$\Rightarrow \frac{-\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B}_0 \times \ell \times \mathrm{y}\right)=\mathrm{BV} \ell$
$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{B}(\hat{\mathrm{i}})(\ell)(-\hat{\mathrm{j}}) \\ & \mathrm{ma}=-\mathrm{B}_0\left[\frac{\mathrm{~B}_0 \mathrm{~V} \ell}{\mathrm{R}}\right](\ell) \\ & \mathrm{a}=-\frac{\mathrm{B}_0^2 \ell^2 \mathrm{~V}}{\mathrm{mR}}\end{aligned}$ $\begin{aligned} & \text { Also } K=\frac{B_0^2 \ell^2 V}{R M} \\ & \text { So }[a=-k v] \\ & \frac{d v}{d t}=-k v \\ & \int_{v_0}^v \frac{d v}{d t}=\int_0^t-k d t \\ & \ell n \frac{v}{v_0}=-k t\end{aligned}$ $\left[\mathrm{v}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}}\right]$ $\left[\mathrm{v}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}}\right]$ $\ldots .(\mathrm{i})$ $\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}} \quad(\mathrm{x} \leq \ell)$ $\begin{aligned} & \int_0^{\mathrm{x}} \mathrm{dx}=\int_0^{\mathrm{t}} \mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}} \mathrm{dt} \\ & =\frac{\mathrm{v}_0}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)\end{aligned}$ When $x=\ell$ $\ell=\frac{\mathrm{V}_0}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)$ Option (D) $\left(\mathrm{v}_0=3 \mathrm{k} \ell\right)$ $\begin{aligned} & \ell=\frac{3 \mathrm{k} \ell}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right) \\ & \frac{1}{3}=1-\mathrm{e}^{-\mathrm{kt}} \\ & \mathrm{f} \frac{2}{3}=2 \mathrm{e}^{-\mathrm{kt}} \\ & -\mathrm{kt}=8 \mathrm{n}\left(\frac{2}{3}\right) \\ & \mathrm{t}=\frac{1}{\mathrm{k}} \ell \mathrm{n}\left(\frac{2}{3}\right)\end{aligned}$ Complete loop will enter at $\mathrm{t}=\frac{1}{\mathrm{k}} \ln \left(\frac{2}{3}\right)$ Option (B) $\frac{\mathrm{d} \phi}{\mathrm{dt}}=0, \underline{\underline{\mathrm{e}}}=0, \mathrm{i}=0, \mathrm{~F}=0$ Ans. B,D)

Hence, the correct answer is options: 'B', 'D'.