There is a conducting ring of radius `R`. Another ring having current `i` and radius `r (r lt lt R)` is kept on the axis of bigger ring such that its center lies on the axis of bigger ring at a distance `x` from the center of bigger ring and its plane is perpendicular to that axis. The mutual inductance of the bigger ring due to the smaller ring is

Options

A. `(mu_(0)piR^(2)r^(2))/((R^(2) + x^(2))^(3//2))`
B. `(mu_(0)piR^(2)r^(2))/(4(R^(2) + x^(2))^(3//2))`
C. `(mu_(0)piR^(2)r^(2))/(16(R^(2) + x^(2))^(3//2))`
D. `(mu_(0)piR^(2)r^(2))/(2(R^(2) + x^(2))^(3//2))`

Answer

D

Explanation

Let current `i` flows in the bigger ring, then the magnetic field on its axis
`B = (mu_(0)iR^(2))/(2(R^(2) + x^(2))^(3//2))`
Flux linked with the smaller ring : `phi = Bpi r^(2)`
`phi = (mu_(0) iR^(2))/(2(R^(2) + x^(2))^(3//2)) pir^(2) = Mi`
`:. M = (mu_(0)piR^(2)r^(2))/(2(R^(2) + x^(2))^(3//2))`