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
$e=-\frac{d\varphi }{dt}\dots \dots (i)$
and the magnetic flux is given as
$\varphi =BA\cos \theta$
(Here, $\theta$ is the angle between the magnetic field $B$ and area vector of ring $A$ )
So,$\theta =0^{\circ }\Rightarrow \varphi =BA\cos 0^{\circ }\Rightarrow \varphi =BA...(ii)$
From Eqs. (i) and (ii), we get
$e=-\frac{d}{dt}(BA)$
So, $e=-\frac{d}{dt}\left[\left(B_0+\alpha t\right)A\right]($ given $;B=B_0+\alpha t)$
$\Rightarrow =-A\frac{d}{dt}\left[B_0+\alpha t\right]$
$\Rightarrow =-A[0+\alpha ]=-A\alpha$
But $A=\pi r^2$ is the area of ring, here $r$ is the radius of ring.
So, $e=-\pi r^2\alpha$
$\Rightarrow e=-\pi \alpha r^2$