Question

Magnetic field in the cylindrical region with its axis passing through $O$ varies at a constant rate $\frac{dB}{dt}.$ A triangular imaginary loop $ABC,$ with $AB=BC,$ is lying in this region as shown in the adjacent figure. The work done to move unit positive charge from $A$ to $B$ along the side $AB$ is:

• A. $r^2\frac{dB}{dt}$
• B. $\pi r^2\frac{dB}{dt}$
• C. $\frac{r^2}{2}\frac{dB}{dt}$
• D. $\frac{3r^2}{2}\frac{dB}{dt}$

Answer 1

The correct option is C $\frac{r^2}{2}\frac{dB}{dt}$

Solution

Draw on another isosceles triangle

similar to $△ABC$ about the side

AC. So,

$W_{AB}=\frac{1}{4}{W}_{ABCD}$

WABCD = work done to

move a unit the change in

the loop. so,

wABCD =$\epsilon mt$

By Faraday's Law,

$\epsilon mt=\frac{d\varphi }{dt}=\frac{dB\left(B{r}^2\right)}{dt}$

$\epsilon mt=\frac{\left(2r^2\right)dB}{dt}$

WAB = $\frac{1}{4}\epsilon mt=\frac{1}{4}\times 2^{r^2}\frac{dB}{dt}=\left(\frac{r^2}{2}\right)\frac{dB}{dt}$

option - C is correct.