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. 3 J
B. `sqrt3J`
C. `(3)/(2)J`
D. `2sqrt3J`

Answer

A

Explanation

`W=MB(cos theta_(1)-cos theta_(2))`
`sqrt3=MB(cos 0^(@)-cos 60^(@))`
`sqrt3=(MB)/(2)" ..(1)"`
`tau=MB sin theta`
`tau = MB sin 60^(@)=sqrt3(MB)/(2)" …(2)"`
`therefore tau=(sqrt3)(sqrt3)=3J`