Question

The magnetic flux density $B$ is changing its magnitude at a constant rate $dB/dt$. A given mass $m$ of copper having resistivity $\rho$ and density $d$, drawn into a wire of radius $a$ and formed into a circular loop of radius $r$ is placed perpendicular to the field $B$. The induced current in the loop $i$ is :

• A. $\frac{m}{2\pi \rho d}\frac{dB}{dt}$
• B. $\frac{m}{4\pi a^{2}r}\frac{dB}{dt}$
• C. $\frac{m}{4\pi ad}\frac{dB}{dt}$
• D. $\frac{m}{4\pi \rho d}\frac{dB}{dt}$

Answer 1

The correct option is D $\frac{m}{4\pi \rho d}\frac{dB}{dt}$

Area of loop, $A=\pi r^2$

Magnetic flux passing through coil is

$\varphi =BA\cos \theta =BA\cos 0^{\circ }=B\left(\pi r^2\right)$

From Faraday's law of induction

$E=\frac{d\varphi }{dt}=\frac{d}{dt}B\left(\pi r^2\right)$

$\therefore E=\pi r^2\frac{dB}{dt}$

Length of the loop, $l=2\pi r$
and area of cross-section of wire, $A^{\prime }=\pi a^2$

So, resistance of the wire will be

$R=\frac{\rho l}{A^{\prime }}=\frac{\rho \left(2\pi r\right)}{\pi a^2}$

Induced current in the loop will be

$i=\frac{E}{R}=\frac{\pi r^2\frac{dB}{dt}}{\frac{\rho \left(2\pi r\right)}{\pi a^2}}=\frac{\pi a^{2}r}{2\rho }\frac{dB}{dt}.........\left(1\right)$

Further mass of wire is,

$m=\left(\pi a^2\right)\left(2\pi r\right)\left(d\right)$

$\Rightarrow \pi a^{2}r=\frac{m}{2\pi d}$

Putting in equation $\left(1\right)$, we get

$i=\frac{m}{4\pi \rho d}\frac{dB}{dt}$

Hence, option $\left(D\right)$ is correct.