91Ó°ÊÓ

The magnitude of a magnetic dipole moment is, $\mathrm{μ}=0.020J/T$

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

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

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:

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

We have the equation for potential energy of dipole moment as

$\begin{array}{rcl}U& =& -\mathrm{μ}B\cos \mathrm{θ}\end{array}$

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

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

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

$K=-\mathrm{μ}B\cos {\mathrm{θ}}_i+1$

Rearranging this equation for angle, we get

$\begin{array}{rcl}{\mathrm{θ}}_i& =& {\cos }^{-1}\left(1-\frac{K}{\mathrm{μ}B}\right)\\ {\mathrm{θ}}_i& =& {\cos }^{-1}\left(1-\frac{8.00×{10}^{-4}}{0.020×5.2×{10}^{-2}}\right)\\ {\mathrm{θ}}_i& =& 76.7^o\\ {\mathrm{θ}}_i& ≈& {77}^o\end{array}$

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

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,${\mathrm{θ}}_f={77}^o$