Question

A small circular loop of radius $l=1cm$ is kept inside a larger circular loop of radius r such that both are coplanar and concentric. If the current in larger loop is as shown in the graph, then current induced in the smaller loop any time is represented as $I_n$ at any instant. Take anticlockwise current as positive. List -1 contains time instants, and List-2 contains the values for $I_n$. Match them. Take $\frac{{\mu }_0\pi }{2Rr}=1$.

$\begin{array}{cccc}& \text{List -1}& & \text{List -2}\\ \left(I\right)& 0sec& \left(p\right)& 25\mu A\\ \left(II\right)& \frac{1}{3}sec& \left(Q\right)& 12.5\mu A\\ \left(III\right)& \frac{3}{2}sec& \left(R\right)& 0\mu A\\ \left(IV\right)& \frac{5}{3}sec& \left(S\right)& -6.25\mu A\\ & & \left(T\right)& -12.5\mu A\\ & & \left(U\right)& -25\mu A\end{array}$

• A. $I\to P,II\to R,III\to U,IV\to P$
• B. $I\to U,II\to R,III\to P,IV\to U$
• C. $I\to P,II\to R,III\to P,IV\to P$
• D. $I\to U,II\to R,III\to U,IV\to U$

Answer 1

The correct option is D $I\to U,II\to R,III\to U,IV\to U$

$\varphi =BA=\frac{{\mu }_{0}i}{2r}\pi l^2$
$\in =\frac{d\varphi }{dt}$
$\Rightarrow \in =\frac{{\mu }_0\pi }{2r}{l}^2\left(\frac{di}{dt}\right)$
$I_n=\frac{\in }{R}$
$\Rightarrow I_n=\frac{{\mu }_0\pi }{2rR}{l}^2\left(\frac{di}{dt}\right)={10}^{-4}\left(\frac{di}{dt}\right)=100\left(\frac{di}{dt}\right)\mu A$
at $t=0,\frac{di}{dt}=\frac{\frac{1}{16}}{\frac{1}{4}}=\frac{1}{4}\Rightarrow I_n=25\mu A$ clockwise
So at $t=0,I_n=-25\mu A$
at $t=\frac{1}{3}sec,\frac{di}{dt}=0\Rightarrow I_n=0$
at $t=\frac{3}{2}sec,\frac{di}{dt}=\frac{1}{4}\Rightarrow I_n=25\mu A$ clockwise
At $t=\frac{3}{2},I_n=-25\mu A$
at $t=\frac{5}{3}sec,\frac{di}{dt}=\frac{1}{4}\Rightarrow I_n=25\mu A$ clockwise.
$\therefore$ At $t=\frac{5}{3},I_n=-25\mu A$
So, $I\to U,II\to R,III\to U,IV\to U$