A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of `60^(@)` and the work done is W. The torque on the magnetic needle at this position is

Options

A. `2sqrt3W`
B. `sqrt3W`
C. `(sqrt3)/2W`
D. `(sqrt3)/4W`

Answer

B

Explanation

Work done, `W = MB (costheta_(1) - costheta_(2))`
where M is the magnetic moment of the magnetic needle.
Here, `theta_(1) = 0^(@),theta_(2) = theta = 60^(@)`
`therefore W = MB(cos0^(@) - cos60^(@))`
`W = MB (1-1/2)=(MB)/2` .....(i)
Torque on the needle is `vectau=vecM xx vecB`
In magnitude, `tau = MBsintheta`
`tau = MB sin 60^(@) = sqrt3/2MB` .....(ii)
Dividing (ii) by (i), we get
`tau/W = sqrt3/2 MB xx 2/(MB) = sqrt3 or tau = sqrt3W`