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 `2xx10^(-4) m^2` and 1000 turns, but of the same magnetic moment . Determine the current flowing through the solenoid.

Explanation

a. `tau=mBsintheta,theta=30^@`, hence `sintheta=1/2`
Thus, `0.016=mxx(800xx10^(-4)T)xx(1/2)`
`m=160xx2/800=0.40 Am^2`
b. The most stable position is `theta =0^@` and the most unstable position is `theta = 180^@ ` . Work done is given by `W=U_m(theta=180^@)-U_m(theta=0^@)^2mB=2xx0.40xx800xx10^(-4) =0.064J`
c. `m_s=N` [A. From part (a), `m_s=0.40Am^2` )
`0.40 = 1000xxIxx2xx10^(-4)`
`I=(0.40xx10^4)/((1000xx2))=2A`