A magnetic needle lying parallel to a magnetic field requires `W units` of work to turn it through `60^(@)`. The torque needed to maintain the needle in this position will be

Explanation

Work done by the magnetic field,
`W=MB(costheta_(1)-costheta_(2))` Here `theta_(1)=60^(@)` and `theta_(2)=0^(@)`
`therefore W=20xx5xx10^(-4)[cos60^(@)-cos0^(@)]`
`=10^(-2)[(1)/(2)-1]=-5xx10^(-3)J`.