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Question: A current-carrying circular coil of magnetic moment \( M \) is situated in a magnetic field \( B \) ...

A current-carrying circular coil of magnetic moment MM is situated in a magnetic field BB . The work done in deflecting it from an angle 00^\circ to θ\theta ^\circ will be :
A) MBMB
B)   MB(1cosθ)\;MB\left( {1 - cos\theta } \right)
C) MB- MB
D) MB(1sinθ)MB\left( {1 - sin\theta } \right)

Explanation

Solution

The current-carrying coil acts as a bar magnet placed in an external magnetic field. The potential energy of a current-carrying coil placed in an external magnetic field depends on the angle of orientation of the current-carrying coil with the external magnetic field. The work done in deflecting the current-carrying coil will be calculated as the difference of the potential energy of the current-carrying coil.

Formula used: In this solution, we will use the following formula:
U=MBU = - M \cdot B where UU is the potential energy stored in the apparent magnet formed by the current-carrying coil and BB is the magnitude of the external magnetic field.

Complete step by step answer:
We’ve been given that a current-carrying coil is placed in an external magnetic field. We know that a current-carrying coil generates a magnetic field of its own which implies that the current-carrying coil can be treated as a bar magnet of magnetic moment MM placed in an external magnetic field.
The potential energy associated with this configuration when the current-carrying coil and the external magnetic field have an angle θ\theta between them will be
U=MB=MBcosθU = - M \cdot B = - MB\cos \theta
Then the work done in deflecting the coil from an angle θ1{\theta _1} to θ2{\theta _2} will be due to the difference of the potential energies of the configuration in the two angles. So, we have
W=U2U1W = {U_2} - {U_1}
W=MBcosθ2(MBcosθ1)\Rightarrow W = - MB\cos {\theta _2} - ( - MB\cos {\theta _1})
Substituting the value of θ1=0{\theta _1} = 0^\circ and θ2=θ{\theta _2} = \theta ^\circ , we get
W=MBcosθ+MBW = - MB\cos \theta + MB
W=MB(1cosθ)\Rightarrow W = MB(1 - \cos \theta ) which corresponds to option (B).

Note:
The reason for the fact that the change in potential energy will be equal to the work done is that there is no external force acting on the coil and the only change in energy of the coil will due to a change in its alignment with the external magnetic field that is a change in its stored potential energy. The coil will not have kinetic energy since it will only rotate in the magnetic field and not translate in any direction.