Question

A conducting loop of resistance $r$ is placed in a non-uniform magnetic field of strength $B=B_0{t}^3$ perpendicular to the plane of the loop. Find the charge flown in the loop in the time interval $t_0$.

• A. $\frac{B_0{t}_0^3\pi R^2}{r}$
• B. $\frac{B_0{t}_0^2\pi R^2}{2r}$
• C. $\frac{2B_0{t}_0\pi R^2}{r}$
• D. $\frac{B_0^2{t}_0\pi R^2}{2r}$

Answer 1

The correct option is A $\frac{B_0{t}_0^3\pi R^2}{r}$

Flux linked with the coil is,

$\varphi =\overrightarrow{B}\cdot \overrightarrow{A}=|\overrightarrow{B}||\overrightarrow{A}|\cos \theta$

$=B_0{t}^3\pi R^2\cos 0^{\circ }=B_0{t}^3\pi R^2$

Thus, the magnitude of the induced $\text{emf}$ in the loop is,

$\left|\begin{array}{c}E\end{array}\right|=|-\frac{d\varphi }{dt}|=\frac{d}{dt}\left(B_0{t}^3\pi R^2\right)=3B_0{t}^2\pi R^2$

Hence, current in the loop is,

$i=\frac{|E|}{r}=\frac{dq}{dt}$

$\Rightarrow dq=\frac{|E|}{r}dt$

If $q$ is the charge flow in the time interval $t_0$, then

${\int }_0^{q}dq={\int }_{t=0}^{t=t_0}\frac{3B_0\pi R^2{t}^2}{r}dt$

$q-0=\frac{3B_0\pi R^2}{r}[\frac{t^3}{3}{]}_0^{t_0}$

$\therefore q=\frac{B_0{t}_0^3\pi R^2}{r}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.