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A short bar magnet is placed in an exter...

A short bar magnet is placed in an external magnetic field of 600 G. When its axis makes an angle of 30° with the external field, it experiences a torque of 0.012 Nm. What is the magnetic moment of the magnet ?

A

`0.2 "Am"^(2)`

B

`0.3 "Am"^(2)`

C

`0.4 "Am"^(2)`

D

`0.6 "Am"^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`tau - mB sin theta `
`therefore m- (tau)/( B sin theta) = (0.012)/(600 xx 10^(-4) xx sin30^(@) )`
`therefore m= (0.012)/(6 xx 10^(-2) xx (1)/(2) ) therefore m= 0.4 "Am"^(2)`
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