Question

A conducting loop of radius $a$ and resistance $R$ is placed in uniform magnetic field perpendicular to its plane. If the bop is rotated by ${180}^{\circ }$ about its diameter then the amount of charge flown through the bop will be.

• A. $\frac{{\pi a}^{2}B}{R}$
• B. $\frac{2{\pi a}^{2}B}{R}$
• C. $\frac{\pi a^{2}B}{2R}$
• D. zero

Answer 1

Answer: (A)

$\frac{{\pi a}^{2}B}{R}$

Given: radius of loop$=a$

resistance of loop$=R$

loop rotated by ${180}^{\circ }.$ $.....\left(1\right)$

To find: amount of charge flow,$q=?$

Solution: Let time period of loop is $T.$

On using $e{q}^n\left(1\right),$ we conclude that time taken to rotate by

${180}^{\circ }$ is $t=T/2.$ $..\left(1\right)$

Now, magnetic flux,$\varphi =BAcos\left(\omega t\right)$

here $A$ is the area of loop and $B$ is magnetic field.

$\varphi =B\pi a^{2}cos\left(\omega t\right)$

now e.m.f induced,$\xi =-\frac{d\varphi }{dt}$

$==>\xi =-\frac{d}{dt}B\pi a^{2}cos\left(\omega t\right)$

$\xi =B\pi a^2\omega sin\left(\omega t\right)$

also,$\xi =IR=R\frac{dq}{dt}$

we get,

$R\frac{dq}{dt}=B\pi a^2\omega sin\left(\omega t\right)$

integrating on both sides w.r.t. $t$,we get

$==>Rq=-B\pi a^{2}cos\left(\omega t\right)$

now $use$ $e{q}^n\left(1\right)$

$==>Rq=-B\pi a^{2}cos\left(\frac{2\pi T}{T\ast 2}\right)$

$==>Rq=-B\pi a^{2}cos\pi$

$==>Rq=B\pi a^2$

$==>q=\frac{B\pi a^2}{R}$

hence,

The correct opt: A