The work done in moving a dipole from its most stable to most unstable position in a `0.09 T` uniform magnetic field is (dipole moment of this dipole = `0.5 Am^(2))`

Options

A. `0.07J`
B. `0.08J`
C. `0.09J`
D. `0.1J`

Answer

C

Explanation

Since the most stable position is at `theta=0` and the most unstable position is at `theta=180^(@)`, then the work done is given by
`w=underset(theta=0^(@))overset(theta=180^(@))inttau(theta)d theta=underset(theta=0^(@))overset(theta=180^(@))int mB sin theta d theta=-mB[cos theta]_(0)^(180^(@))`
`=-mB[cos 180^(@)-cos 0^(@)]=-mB[-1-1]`
`=-mB[-2]=2mB`
`therefore W=2xx0.50xx0.09=0.09J`