Permanent magnetismEasy
Question
A short bar magnet placed with its axis at 30o with a uniform external magnetic field of 0.16T experiences a torque of magnitude 0.032 J. The magnetic moment of he bar moment will be
Options
A.0.23 J/T
B.0.40 J/T
C.0.80J/T
D.Zero
Solution
MB sinθ = T ⇒ M × 0.16 × 
= 0.032 ⇒ M = 0.4 J/T
= 0.032 ⇒ M = 0.4 J/TCreate a free account to view solution
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