A short bar magnet placed with its axis at 30 with an external field of 800 Gauss, experiences a torque of 0.016 N m. The work done in moving it from most stable to most unstable position is 10-3 J. The value of is .
Answer
64
Solution
B = 800 G = 0.08 T
Torque: = mB 0.016 = m 0.08 30
m = 0.4 A m2
Most stable: = 0 (U = -mB); most unstable: = 180 (U = +mB)
W = Uf - Ui = 2mB = 2 0.4 0.08 = 0.064 J = 64 10-3 J
= 64
Answer: 64