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The magnetic field on the axis of a curr...

The magnetic field on the axis of a current carrying circular coil of radius a at a distance 2a from its centre will be:

A

`(mu_(0)lR^(2))/(2(R^(2)+r^(2))^((3)/(2)))`

B

`(mu_(0)lr^(2))/(2(R^(2)+r^(2))^((3)/(2)))`

C

`(mu_(0)l)/(2r)`

D

`(mu_(0)l)/(2R)`

Text Solution

Verified by Experts

The correct Answer is:
B
A coil of radius e carrying I, magnetic field at a point a distance R from the centre of the coil is given by
`B = (mu_(0)Ir^(2))/(2[R^(2)+r^(2)]^(3//2))`
At the centre of a coil, R = 0
`therefore B = (mu_(0)Ir^(2))/(2r^(3))=(mu_(0)I)/(2r^(3))`
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