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

Explanation

Here, `theta_1=0^@`, `theta_2=60^@`, `W=sqrt3J`, `tau=?`
As `W=-MB(cos theta_2-cos theta_1)`
`:. sqrt3=-MB(cos60^@-cos0^@)`
`=-MB(1/2-1)`
or `MB=2sqrt3`
`tau=MBsin theta=2sqrt3sin 60^@`
`=2sqrt3xxsqrt3/2=3J`