Question

A loop shown in the figure is immersed in the varying magnetic field $B=B_{0}t$, directed into the page. If the total resistance of the loop is $R$, then the direction and magnitude of induced current in the inner circle is :

• A. clockwise $\frac{B_0(\pi a^2-b^2)}{R}$
• B. anticlockwise $\frac{B_0(\pi a^2+b^2)}{R}$
• C. clockwise $\frac{B_0(\pi a^2+{4b}^2)}{R}$
• D. clockwise $\frac{B_0({4b}^2-\pi a^2)}{R}$

Answer 1

The correct option is D clockwise $\frac{B_0({4b}^2-\pi a^2)}{R}$

Surface area of the loop perpendicular to the field is the difference of surface area of square loop and the circular loop.

Hence, $S=(2b{)}^2-\pi a^2$

$S=4b^2-\pi a^2$

$\left|e\right|=\frac{d\varphi }{dt}=S\cdot \frac{dB}{dt}$$=(4b^2-\pi a^2)B_0$

$i=\frac{\left|e\right|}{R}=\frac{(4b^2-\pi a^2)B_0}{R}$

$⨂$ magnetic field is increasing. So,

$⨀$ magnetic field is produced.