Permanent magnetismHard
Question
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60. The torque needed to maintain the needle in this position will be
Options
A.
B.2√3 J
C.3 J
D.√3 J
Solution
Work done Uf - Ui = - MBcos60o - (-MBcos0o)
= - MB
MB = 2√3
= MB sin 60o = 
= - MB
MB = 2√3 = MB sin 60o = Create a free account to view solution
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