Question

The magnetic field varies with time as $B=0.5t^2$, in a square coil of side $10\text{cm}$, kept with its plane perpendicular to the field. If the resistance of the coil is $5\times {10}^{-2}\Omega ,$ the total charge that flows through it from $t=0$ to $t=2\text{sec}$ will be:

• A. $1\text{C}$
• B. $0.1\text{C}$
• C. $0.2\text{C}$
• D. $0.4\text{C}$

Answer 1

The correct option is D $0.4\text{C}$

Given,

$B=0.5t^2$

side of the coil $\left(a\right)=10\text{cm}=0.1\text{m}$

Angle between $\overrightarrow{B}$ and $\overrightarrow{A}$, $\theta =0^{\circ }$

Resistance of the coil $\left(R\right)=5\times {10}^{-2}\Omega$

$t_1=0\text{sec}$

$t_2=2\text{sec}$

Flux linked with the coil is given by,

$\varphi =BA\cos \theta =0.5t^2\times (0.1{)}^2\times \cos 0^{\circ }$

$\Rightarrow \varphi =5\times {10}^{-3}{t}^2$

$\therefore d\varphi ={10}^{-2}tdt$

Now,

$E=\left|\frac{d\varphi }{dt}\right|$

$IR=\left|\frac{d\varphi }{dt}\right|$

$\frac{dq}{dt}R=\left|\frac{d\varphi }{dt}\right|$

$\therefore dq=\frac{1}{R}\left|d\varphi \right|=\frac{1}{5\times {10}^{-2}}\times {10}^{-2}tdt$

$\therefore q=\int dq={\int }_0^2\frac{tdt}{5}$

$q=\frac{1}{5}{\left[\frac{t^2}{2}\right]}_0^2=\frac{1}{5}\left(\frac{(2{)}^2-0}{2}\right)=0.4\text{C}$

Hence, option ($D$) is the correct answer.