A magnetic dipole with a dipole moment of magnitude 0.020 J/T is released from rest in a uniform magnetic field of magnitude 52 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.80mT. (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

The magnitude of a magnetic dipole moment is, $\mu =0.020J/T$

The magnetic field is, $B=52mT=5.2\times {10}^{-2}T$

Kinetic energy of dipole moment is,$K=0.80mJ=8.00\times {10}^{-4}J$

Here, we need to use the equation of principle of conservation of energy and the equation of potential energy of dipole. We can solve these equations simultaneously to get the required angles.

Formulae:

$\begin{array}{rcl}{K}_i+U_i& =& K_f+U_f\\ U& =& -\mu B\cos \theta \end{array}$

We have the equation for potential energy of dipole moment as

$\begin{array}{rcl}U& =& -\mu B\cos \theta \end{array}$

Now, according to the conservation of energy principle,

$K_i+U_i=K_f+U_f$

Assuming the initial kinetic energy of a dipole moment as zero, we get

$$\begin{gathered} K_f=K=U_i-U_f \\ K=-\mu B\cos {\theta }_i-\left(-\mu B\cos {\theta }_f\right) \end{gathered}$$

For the final state, the magnetic dipole is in the direction of magnetic field, so ${\theta }_f=0^o$.

$K=-\mu B\cos {\theta }_i+1$

Rearranging this equation for angle, we get

$\begin{array}{rcl}{\theta }_i& =& {\cos }^{-1}\left(1-\frac{K}{\mu B}\right)\\ {\theta }_i& =& {\cos }^{-1}\left(1-\frac{8.00\times {10}^{-4}}{0.020\times 5.2\times {10}^{-2}}\right)\\ {\theta }_i& =& 76.7^o\\ {\theta }_i& \approx & {77}^o\end{array}$

Hence$\begin{array}{rcl}{\theta }_i& =& {77}^o\end{array}$

Since we are assuming that energy is conserved, the dipole will oscillate like a simple pendulum.

So, it will be momentarily at rest at the same angle on the opposite side.

Hence,${\theta }_f={77}^o$