When a bar magnet (magnetic dipole) of dipole moment \(\vec B_m\) is held at an angle 0 with the direction of a uniform magnetic field B , the magnitude of the torque acting on the dipole is
If the dipole is rotated through a very small angular displacement dθ against the torque τB at constant angular velocity, then the work done by external torque (\(\vec{\tau}_{ext}\)) for this small angular displacement is given by
dW = (\(\vec{\tau}_{ext}\)) dθ …… (2)
The bar magnet has to be moved at constant angular velocity, which implies that
\(|\vec{\tau}_B|\) = |\(\vec{\tau}_{ext}\)|
dw = Pm B sin θ d θ
Total work done in rotating the dipole from θ’ to θ is
W = -Pm B (cosθ – cosθ’) …… (3)
This work done is stored as potential energy in bar magnet at an angle θ when it is rotated from θ’ to θ and it can be written as
U = -Pm B (cosθ – cosθ’) …(4)
In fact, the equation (4) gives the difference in potential energy between the angular positions θ’ and θ. We can choose the reference point θ’ = 90° , so that second term in the equation becomes zero and the equation 4 can be written as
The potential energy stored in a bar magnet in a uniform magnetic field is given by
