Question

A conducting circular loop of face area $2.5\times {10}^{-3}{\text{m}}^2$ is arranged with its plane perpendicular to a magnetic field which varies as $B=0.20\sin \left(50\pi t\right)$. Find the net charge flowing through the loop during the interval $t=0$ to $t=40\text{ms}$.

Assume resistance of the loop to be $10\Omega$.

• A. $1\text{C}$
• B. $0.5\text{C}$
• C. $0.25\text{C}$
• D. Zero

Answer 1

The correct option is D Zero

The flux through the loop at any time $t$ is given by,

$\varphi =B_{0}A\sin \omega t$

$\therefore$ The emf induced in the loop is,

$E=\left|\frac{d\varphi }{dt}\right|=\left|B_{0}A\frac{d}{dt}\right(\sin \omega t\left)\right|$

$E=B_{0}A\omega \cos \omega t$

The induced current is given by,

$i=\frac{E}{R}=\frac{B_{0}A\omega }{R}\cos \omega t$

$\Rightarrow i=\frac{0.2\times 2.5\times {10}^{-3}\times 50\pi }{10}\cos \omega t$

$i=7.85\times {10}^{-3}\cos \omega t$

$\therefore$ Current changes sinusoidally with the time.

The time period is given by,

$T=\frac{2\pi }{\omega }=\frac{2\pi }{50\pi }=40\text{ms}$

The charge flowing through any cross - section in $40\text{ms}$ is,

$Q={\int }_0^{T}idt$

$Q={\int }_0^{T}7.85\times {10}^{-3}\cos \omega tdt$

$Q=-\frac{7.85\times {10}^{-3}}{50\pi }[\sin \omega t{]}_0^T$

Now,

$[\sin \omega t{]}_0^T=(\sin \omega T-\sin 0^0)$

$=\sin (\frac{2\pi }{T}\times T)-\sin 0^0=0$

$\therefore Q=0$

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