Question

The magnetic field $B = 2t^2 + 4t^2$ (where t = time) is applied perpendicular to the plane of a circular wire of radius $r$ and resistance $R$. If all the units are in $SI$ the electric charge that flows through the circular wire during $t = 0 s$ to $t = 2 s$ is

• A. $\frac{6\pi\,r^{2}}{R}$
• B. $\frac{20\pi\,r^{2}}{R}$
• C. $\frac{32\pi\,r^{2}}{R}$
• D. $\frac{48\pi\,r^{2}}{R}$

Answer 1

Answer: (B)

Given, $B=2 \,t+4 \,t^{2}$
at $t=0, B_{1}=0$
and at $=2, B_{2}=2 \times 2+4(2)^{2} $
$=4+16=20 \,Wb / m ^{2} $
We have, $\Delta Q =\frac{\Delta \phi}{R}=\frac{\pi r^{2}\left(B_{2}-B_{1}\right)}{R} $
$ \therefore =\frac{\pi r^{2}[20-0]}{R}=\frac{20 \pi r^{2}}{R} $