A magnetic needle lying parallel to a magnetic field needs `2J` work to turn it through `60^@`. Calculate the torque required to maintain the needle in this position.

Answer

`3*46N-m`

Explanation

Here, `theta_1=0^@`, `W=2J`, `theta_2=60^@`, `tau=?`
`W=-MB(costheta_2-costheta_1)`
`2=-MB(cos60^@-cos0^@)`
`=-MB(1/2-1)=(MB)/(2)`
`MB=2xx2=4`
Torque required, `tau=MBsintheta`
`=4sin60^@=4xxsqrt3/2=3*46Nm`