A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet ?
(b) What is the work done in moving it from its most stable to most unstable position ?
(c) The bar magnet is replaced by a solenoid of cross-sectional area `2 xx 10^(-4) m^(2)` and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid.

Explanation

(a) Torque exerted on magnetic dipole (here short bar magnet) is,
`tau = m B sin theta`
`therefore m= (tau)/( B sin theta)`
`therefore m = (0.016)/( 800 xx 10^(-4) xx sin 30^@)`
`therefore m= (0.016)/( 0.08 xx 0.5)`
`therefore = 0.4` (ampere) (meter)`""^(2)`
(b) For most stable state `overset(to) (m) || overset(to) (B) rArr theta_1 = 0^@`
For most unstable state `overset(to) (m) || (-overset(to) (B)) rArr theta_2 = 180^@`
Now required amount of work is,
`W = Delta U` (Change in magnetostatic potential energy)
`= U_2 = U_1`
`= - m B cos theta_2 - (- m B cos theta_1)`
`= m B cos theta_1 - m B cos theta_2`
`therefore W= m B (cos theta_1 - cos theta_2)`
`= m B ( cos 0^@ - cos 180^@)`
`= m B {1-(-1)}`
`therefore W= 2 m B`
`therefore W= (2) (0.4) (800 xx 10^(-4) )`
`therefore W= 0.064` J
(c) We know that current carrying solenoid behaves like a bar magnet with magnetic dipole moment,
`m_s = NIA`
`therefore m= NIA (because m_s = m` is given `)`
`therefore I= (m)/( NA)`
`therefore I= (0.4)/( 1000 xx 2 xx 10^(-4) ) `
`therefore I= 2A`