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) From Eq. (5.3). `tau =m B sin theta, theta= 30^@,` hence `sin theta=1//2.`
Thus, `0.016 = m xx (800 xx 10^(-4)T) xx (1//2)`
`m =160 xx 2//800 = 0.40Am^(2)`
(b) From Eq. (5.6). the most stable position is `0 = 0^@` and the most unstable position is `theta= 180^@.` Work done is given by
`W=U_(m) (theta-180^(@))-U_(m) (theta=0^(@))`
`=2mB =2 xx 0.40 xx 800 xx 10^(-4)=0.064J`
(c) From Eq. (4.30) `m_(s)=N//A`. From part (a), `m_(s)=0.40 Am^(2)`
`0.40=1000 xx l xx 2 xx 10^(-4)`
`I=0.40 xx 10^(4)//(1000 xx 2)=2A`