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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^2and 1000 turns, but of the same magnetic moment . Determine the current flowing through the solenoid. |
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Answer» Solution :(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` |
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