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:

\(\Rightarrow \frac{-\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B}_0 \times \ell \times \mathrm{y}\right)=\mathrm{BV} \ell\)

\(\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{B}(\hat{\mathrm{i}})(\ell)(-\hat{\mathrm{j}}) \\ & \mathrm{ma}=-\mathrm{B}_0\left[\frac{\mathrm{~B}_0 \mathrm{~V} \ell}{\mathrm{R}}\right](\ell) \\ & \mathrm{a}=-\frac{\mathrm{B}_0^2 \ell^2 \mathrm{~V}}{\mathrm{mR}}\end{aligned}\)
\(\begin{aligned} & \text { Also } K=\frac{B_0^2 \ell^2 V}{R M} \\ & \text { So }[a=-k v] \\ & \frac{d v}{d t}=-k v \\ & \int_{v_0}^v \frac{d v}{d t}=\int_0^t-k d t \\ & \ell n \frac{v}{v_0}=-k t\end{aligned}\)
\(\left[\mathrm{v}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}}\right]\) \(\left[\mathrm{v}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}}\right]\) \(\ldots .(\mathrm{i})\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}} \quad(\mathrm{x} \leq \ell)\)
\(\begin{aligned} & \int_0^{\mathrm{x}} \mathrm{dx}=\int_0^{\mathrm{t}} \mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}} \mathrm{dt} \\ & =\frac{\mathrm{v}_0}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)\end{aligned}\)
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)