A short bar magnet placed with its axis at `30^@` with an extemal field of 800 G experiences a tunque 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 cros B-sectional area `2 xx 10^(-4) m^(2)` and 1000 tums, 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`