Question

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
A. `sqrt(3)W`
B. `W`
C. `sqrt(3)/2W`
D. `2W`

Answer 1

Correct Answer - A
`W=MB(costheta_(1)-costheta_(2))=MB(cos 0^(@)-cos 60^(@))`
`=MB(1-1/2)=(MB)/2`
and `tau=Mbsintheta=MBsin60^(@)=MBsqrt(3)/2`
`:. Tau=((MB)/2)sqrt(3)impliestau=sqrt(3)W`