A magnetic needle suspended parallel to a magnetic field requires `sqrt3J` of work to turn it through `60^@`. The torque needed to maintain the needle in this postion will be:

Options

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

Answer

B

Explanation

In this case, work done
`W=MB(cos theta_(1)-cos theta_(2))`
`=MB(cos 0^(@)-cos 60^(@))=MB(1-1/2)=(MB)/2`
`=MB=2sqrt(3J) ( :' given W=sqrt(3) J)`
`tau=MB sin 60^(@)=(2sqrt(3))(sqrt(3)/2)J=3J`