Question

The radius of the circular conducting loop shown in figure is $R$. Magnetic field is decreasing at a constant rate $\alpha$. Resistance per unit length of the loop is $\rho$. Then, the current in wire $AB$ is ($AB$ is one of the diameters)

• A. $\frac{R\alpha }{2\rho }$ from $A$ to $B$
• B. $\frac{R\alpha }{2\rho }$ from $B$ to $A$
• C. $\frac{R\alpha }{\rho }$ from $B$ to $A$
• D. $0$
• E. $answerrequired$

Answer 1

The correct option is D $0$

emf across AB = $\frac{\pi R^2}{2}\alpha$

In the upper semicircular loop :

resistance of the across AB = $\rho \pi R$

= $\rho \pi R$

current in AB = $\frac{emf}{resistance}=\frac{\pi R^2\alpha }{\rho \pi R2}=\frac{\alpha R}{\rho 2}$ in the clockwise direction.

In the lower semicircular loop:

resistance of the across AB = $\rho \pi R$

= $\rho \pi R$

current in AB = $\frac{emf}{resistance}=\frac{\pi R^2\alpha }{\rho \pi R2}=\frac{\alpha R}{\rho 2}$ in the anticlockwise direction.

Therefore current in AB = 0