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When the magnetic dipole is rotated thro...

When the magnetic dipole is rotated through angle `theta` , work done in rotating dipole is given by `-MB cos theta`. Now, a magnetic needle lying parallel to a magnetic field requires W units of work to turn it through `60^(@)`. What will be the value of torque needed to minimum the needle in same position?

A

`sqrt(3) W`

B

W

C

`(sqrt(3)W)/2`

D

2W

Text Solution

Verified by Experts

The correct Answer is:
A
`WW = MB(cos theta_(1) -cos theta_(2))`
`= MB (cos 0^(@) - cos 60^(@))`
`= MB(1-1/2) =(MB)/2`
and `tau = MB sin theta = MB. Sin 60 = MB sqrt(3)/2`
`therefore tau = (MB)/2 sqrt(3) rArr tau = sqrt(3) W`
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