Question

A small equilateral triangular loop of side $l=1cm$ is kept inside a large circular loop of radius r such that both are coplanar and concentric. If current in the larger loop is as shown in the graph, then find the current induced in the smaller loop during the time instants given. Take anticlock wise current as positive.


List - 1 gives time instants and list - 2 gives induced current in the smaller loop. Take $\frac{{\mu }_0\pi }{2rR}=1$ where R is resistance of small loop.
$\begin{array}{cccc}& \text{List -1}& & \text{List -2}\\ \left(I\right)& 0sec& \left(p\right)& 25\mu A\\ \left(II\right)& \frac{1}{3}sec& \left(Q\right)& 12.5\mu A\\ \left(III\right)& \frac{3}{2}sec& \left(R\right)& 0\mu A\\ \left(IV\right)& \frac{5}{3}sec& \left(S\right)& -6.25\mu A\\ & & \left(T\right)& -12.5\mu A\\ & & \left(U\right)& -25\mu A\end{array}$

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

Answer 1

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

$\Phi =BA=\frac{{\mu }_{0}i}{2r}\frac{\sqrt{3}}{4}{l}^2=\frac{{\mu }_0}{2}\frac{\sqrt{3}}{4}\frac{l^2}{r}\frac{1}{\sqrt{3}}\sin \omega t$
$\in =\frac{d\varphi }{dt}=\frac{{\mu }_0}{8}\frac{l^2}{r}\omega \cos \omega t{I}_{triangle}=\frac{\in }{R}$ and $\omega =\frac{2\pi }{T}=\frac{2\pi }{2}$
$\Rightarrow I_{triangle}=\frac{{\mu }_0{l}^2\pi }{8rR}\cos \omega t=\frac{{\mu }_0\pi }{2rR}.\frac{l^2\cos \omega t}{4}=25\cos \omega t\mu A$
$I_{triangle}=25\cos \omega t\mu A$
at $t=0,I_{triangle}=25\mu A$ clock wise
So at $t=0,I_{triangle}=-25\mu A$
and at $t=\frac{1}{3},I_{triangle}=25\cos \frac{\pi }{3}=12.5\mu A$ clockwise
So $t=\frac{1}{3}\Rightarrow I_{triangle}=-12.5\mu A$
at $t=\frac{3}{2}\Rightarrow I_{triangle}=0$
At $t=\frac{5}{3},I_{triangle}=25\cos \frac{5A}{3}=12.5\mu A$ clockwise
so at $t=\frac{5}{3},I_{triangle}=-12.5\mu A$
$I\to U,II\to T,III\to R,IV\to T$