Question

A small loop of resistance $2\pi \Omega $ and having cross sectional area ${10^{ - 4}}{m^2}$ is placed concentrically and coplanar with a bigger loop of radius $0.1\,m$. A steady current of $1A$ is passed in a bigger loop. The smaller loop is rotated about its diameter with an angular velocity $\omega $ .The value of induced current in the smaller loop will be [in ampere] :-
A. $\pi \times {10^{ - 9}}\omega \sin \omega t$
B. $\dfrac{\pi }{2} \times {10^{ - 11}}\omega \sin \omega t$
C. ${10^{ - 9}}\omega \sin \omega t$
D. ${10^{ - 10}}\omega \sin \omega t$

Answer 1

Hint:To calculate the value of induced current in the smaller loop, we have to use the concept of induced current in which when the magnetic flux passing through a loop changes, an induced emf and hence induced current is produced in the circuit. Also, we use the concept of magnetic field through a circular loop.

Complete step by step answer:
Let us write the data given in the question,
$R = 2\pi \Omega $ , $s = {10^{ - 4}}{m^2}$, $r = 0.1m$ , $I = 1A$
The magnetic field through the circular loop is given by
$B = \dfrac{{{\mu _0}I}}{{2r}}$
Where, $I$ - current flowing through the loop and $r$ -radius of the loop.
$B = \dfrac{{{\mu _0}I}}{{2r}} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 1}}{{2 \times 0.1}}$
$\Rightarrow B = 2\pi \times {10^{ - 6}} - - - - - - - - - (1)$

The electromotive force or induced emf through a loop having a small area $ds$ is given by
$\varepsilon = - \dfrac{{d{\phi _B}}}{{dt}} = - \dfrac{{d\left( {Bs\cos \theta } \right)}}{{dt}}$
$\Rightarrow \varepsilon = - 2\pi \times {10^{ - 6}}\dfrac{{d\bar s}}{{dt}} - - - - - - - - - - (2)$
We have $\bar s = s\cos \omega t$
$\dfrac{{d\bar s}}{{dt}} = - s\omega \sin \omega t$
Substituting in equation $(2)$ , we get
$\varepsilon = - 2\pi \times {10^{ - 6}} \times \left( { - s\omega \sin \omega t} \right)$
$\Rightarrow \varepsilon = 2\pi \omega \sin \omega t \times {10^{ - 6}} \times {10^{ - 4}}$
$\Rightarrow \varepsilon = 2\pi \omega \sin \omega t \times {10^{ - 10}} - - - - - - - (3)$
Now, The induced current through smaller loop is given by
$i = \dfrac{\varepsilon }{R} \\
\Rightarrow i= \dfrac{{2\pi \omega \sin \omega t \times {{10}^{ - 10}}}}{{2\pi }}$
$\therefore i = {10^{ - 10}}\omega \sin \omega t$

Hence, option D is correct.

Note:We should know the magnetic field due to a circular loop and the direction of the induced current is such that the induced emf in the loop due to changing flux always opposes the change in the magnetic flux. When the angle between $\theta $ and $B$ changes by rotating the loop with angular frequency, the magnetic flux changes and emf is induced.