Q. 6

A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is
A. MB

B.

C.

D. zero

Answer :

Potential energy of a loop in a uniform magnetic field is given by


U=-MBcosθ where,


M is magnetic moment of the current carrying loop,


B is magnetic field in which the loop is kept,


θ is the angle between M and B.


So, the change in potential energy of the loop will give the work done by the magnetic field as initial and final kinetic energy is 0. Since, there is no change in angle between M bad B when we are rotating the loop along the axis perpendicular to its plane so, there will be no change in the potential energy of the loop. And, W= Uf – Ui, where Ui is initial potential energy of the loop and Uf is final energy of the loop. So, W=0.


Hence, our answer is option (d).

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