Question

A straight metal wire of length $L$, cross-sectional area $A$ and resistivity $\rho$ is made into a square frame. A uniform magnetic field $B$ is perpendicular to the plane of the frame and is changing at a constant rate $\frac{dB}{dt}$. Find the current induced in the frame.

• A. $\frac{LA}{\rho }\frac{dB}{dt}$
• B. $\frac{LA}{4\rho }\frac{dB}{dt}$
• C. $\frac{LA}{8\rho }\frac{dB}{dt}$
• D. $\frac{LA}{16\rho }\frac{dB}{dt}$

Answer 1

The correct option is D $\frac{LA}{16\rho }\frac{dB}{dt}$

Side of square frame $=\frac{L}{4}$

Area of the frame, $a=\frac{L^2}{16}$

The magnitude of induced emf is given by

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

$\Rightarrow |E|=a\frac{dB}{dt}=\frac{L^2}{16}\frac{dB}{dt}$

Resistance of frame is $R=\frac{\rho L}{A}$

$\therefore$ Induced current,

$i=\frac{E}{R}=\frac{LA}{16\rho }\frac{dB}{dt}$

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