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 $B=0.5\sin \left(100\pi t\right)$. Find the net charge flowing through the loop during the interval $t=0$ to $t=20\text{ms}$.

[Take resistance of the conducting wire as $10\Omega$]

• A. $1\text{C}$
• B. $2\text{C}$
• C. $0.5\text{C}$
• D. $0\text{C}$

Answer 1

The correct option is D $0\text{C}$

The induced current is given by,

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

Substituting the data given in the question gives,

$i=\frac{0.5\times 2.5\times {10}^{-3}\times 100\pi }{10}\cos \omega t$

$i=12.5\pi \times {10}^{-3}\cos \omega t$

$\therefore$ current changes sinusoidally with the time,

The time period is given by,

$T=\frac{2\pi }{\omega }=\frac{2\pi }{100\pi }=20\text{ms}$

The charge flowing through any cross - section in $20\text{ms}$ is,

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

$Q={\int }_0^{T}12.5\pi \times {10}^{-3}\cos \omega tdt$

$Q=-\frac{12.5\pi \times {10}^{-3}}{50\pi }\left[\sin \omega t\right]{|}_0^T$

Now, $[\sin \omega t{]}_0^T=(\sin \omega T-\sin 0)$ $Q=\sin (\frac{2\pi }{T}\times T)-\sin 0=0$

$\therefore Q=0$

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