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There is a conducting ring of radius R. ...

There is a conducting ring of radius R. Another ring of radius `r(r lt lt R)` is kept on the axis of bigger ring such that it’s a center lies on the axis of bigger ring at a distance x from the center of bigger ring and its place is perpendicular to the axis. The mutual inductance of the two rings is

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))`

Text Solution

Verified by Experts

The correct Answer is:
D

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))`
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