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

Here, `B = 600 G = 6000 xx 10^(-4) T 6xx 10^(-2) T`
`theta = 30^@`
`tau = 0.012 Nm`
`N= 1000`
`A= 2 xx 10^(-4) m^(2)`
(a) `tau = m B sin theta`
`therefore m= (tau)/( B sin theta) = (0.012)/( 6 xx 10^(-2) xx sin 30^(@) ) = (0.2)/(½)`
`m= 0.4 "Am"^(2)`
(b) For most stable position `theta_(1) = 0^(@)` and for most unstable position `theta^(2) = 180^@`.
`therefore W= - mB cos theta_(2) - (-mB cos theta_(1) ) `
`=-mB cos 180^(@) + m B cos 0^@`
`=-mB (-1) + mB (1)`
`=mB+ mB`
`= 2mB`
`2 xx 0.4 xx 6 xx 10^(-2)`
`therefore W=0.048` J
(c) Magnetic moment of solenoid,
`m_(s) = NIA, "here" m_(s) = m = 0.4 "Am"^2`
`I= (m)/( NIA) = (0.4)/( 1000 xx 2 xx 10^(-4) ) = 2A`