Question

A conducting ring of radius $r$ is placed perpendicularly inside a time varying magnetic field given by $B=B_{0}+\alpha t .$ $B_{0}$ and $\alpha$ are positive constants. Find the emf produced in the ring.

• A. $-\pi \alpha r^{2}$
• B. $-\pi \alpha r$
• C. $-\pi \alpha^{2} r^{2}$
• D. $-\pi \alpha^{2} r$

Answer 1

Answer: (A)

The induced emf is given by
$e=-\frac{d \phi}{d t} \ldots \ldots (i)$
and the magnetic flux is given as
$\phi=B A \cos \theta$
(Here, $\theta$ is the angle between the magnetic field $B$ and area vector of ring $A$ )
So,$\theta=0^{\circ} \Rightarrow \phi=B A \cos 0^{\circ} \Rightarrow \phi=B A...(ii)$
From Eqs. (i) and (ii), we get
$e=-\frac{d}{d t}(B A)$
So, $e=-\frac{d}{d t}\left[\left(B_{0}+\alpha t\right) A\right]\left(\right.$ given $\left.; B=B_{0}+\alpha t\right)$
$\Rightarrow =-A \frac{d}{d t}\left[B_{0}+\alpha t\right]$
$\Rightarrow =-A[0+\alpha]=-A \alpha $
But $A=\pi r^{2}$ is the area of ring, here $r$ is the radius of ring.
So, $e=-\pi r^{2} \alpha$
$\Rightarrow e=-\pi \alpha r^{2}$