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. `2 sqrt3` J
B. 3 J
C. `sqrt3 J`
D. `(3)/(2) J`

Answer

B

Explanation

`W = Delta U = MB (cos 0^(@) - cos 60^(@))`
`W = (MB)/(2) = sqrt3 implies MB = 2sqrt3 J`
now , `tau = vecM xx vecB = MB sin theta = MB sin 60^(@) = 2 sqrt3 xx (sqrt3)/(2) = 3 J`