Derive an expression for potenstial energy of bar magnet in a uniform magnetic field.

Explanation

When a bar magnet (magnetic dipole ) of dipole moment `vec(P_(m))` is held at an angle `theta` with the direction of a uniform magnetic field B , The magnetitude of the torque acting on the dipole is
` |vec(tau_(B))| = |vec(P_(m))| |vec(B)| sin theta " "` (1)
If the dipole is rotated through a very small angular displacement d`theta` against the torque `tau_(B)` 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))) d theta " " ` ....(2)
the bar magnet has to be moved at constant angular velocity, which implies that
`|vec(tau_(B))| = |vec(tau_(exl))|`
dW = `P_(m) B sin theta d theta`
Total work done in rotating the dipole from `theta`. to `theta` is
`W = int_(theta)^(theta) tau d theta = int_(theta.)^(theta) P_(m) B sin theta d theta = p_(m) B [ - cos theta d theta ]_(theta.)^(theta)`
W = `P_(m) B(cos theta - cos theta.)" "` .... (3) e
This work done is stored as potential energy in ber magnet at an angle `theta` when it is rotated from `theta ` to `theta` and it can be written as ,
` U = -P_(m) B (cos theta - cos theta.) " "` ... (4)
In fact, the equation (4) gives the difference in potential energy between the angular positions `theta. and theta`. we can choose the reference point `theta. = 90^(@)` , so that second term in the equation becomes zero and the equation 4 can be written as
U = `-P_(m) B (cos theta)" "`....(5)
U= -`vec(p)_(m)• vec(B)`
Case 1
(i) If `theta = 0^(@) ,`then
U = `-p_(m) B (cos theta^(@) ) = - p_(m) `B
(ii) `If theta = 180^(@) ,` then
U = -`p_(m) B (cos 180^(@)) = p_(m) `B
The potential energy stored in a bar magnet in a uniform magnetic field is given by