MEC O12

Analysis: We need to calculate the work done by magnetic interactions when a dipole $m_p$ at point P is rotated from an initial orientation to a final orientation.

• Initial Orientation: $\vec{m}_p$ is along the y-axis (parallel to the source dipole $\vec{m}$).
• Final Orientation: $\vec{m}_p$ is aligned along the position vector $\vec{r}$ of point P.

The work done by the magnetic field forces is given by the decrease in potential energy:

$$ W_{mag} = U_{initial} – U_{final} $$

Where potential energy $U = -\vec{m} \cdot \vec{B}$.

1. Decompose the Initial Dipole Vector

The initial dipole $\vec{m}_p$ points in the $\hat{j}$ direction. In polar coordinates $(\hat{e}_r, \hat{e}_{\theta})$:

$$ \hat{j} = \cos\theta \hat{e}_r – \sin\theta \hat{e}_{\theta} $$

So, $\vec{m}_{initial} = m_p (\cos\theta \hat{e}_r – \sin\theta \hat{e}_{\theta})$.

2. Calculate Initial Potential Energy ($U_i$)

Using the field $\vec{B}$ from Question 11: $\vec{B} = \frac{\mu_0 m}{4\pi r^3} [ (2\cos\theta)\hat{e}_r + (\sin\theta)\hat{e}_{\theta} ]$

$$ U_i = – \vec{m}_{initial} \cdot \vec{B} $$ $$ U_i = – [m_p (\cos\theta \hat{e}_r – \sin\theta \hat{e}_{\theta})] \cdot \left[ \frac{\mu_0 m}{4\pi r^3} (2\cos\theta \hat{e}_r + \sin\theta \hat{e}_{\theta}) \right] $$ $$ U_i = – \frac{\mu_0 m m_p}{4\pi r^3} [ 2\cos^2\theta – \sin^2\theta ] $$

Using the identity $\sin^2\theta = 1 – \cos^2\theta$:

$$ 2\cos^2\theta – (1-\cos^2\theta) = 3\cos^2\theta – 1 $$ $$ U_i = – \frac{\mu_0 m m_p}{4\pi r^3} (3\cos^2\theta – 1) $$

3. Calculate Final Potential Energy ($U_f$)

Finally, the dipole aligns with $\vec{r}$, so $\vec{m}_{final} = m_p \hat{e}_r$.

$$ U_f = – [m_p \hat{e}_r] \cdot \vec{B} $$ $$ U_f = – m_p \left( \frac{\mu_0 m}{4\pi r^3} (2\cos\theta) \right) $$ $$ U_f = – \frac{\mu_0 m m_p}{4\pi r^3} (2\cos\theta) $$

4. Calculate Work Done

$$ W = U_i – U_f $$ $$ W = \left[ – \frac{\mu_0 m m_p}{4\pi r^3} (3\cos^2\theta – 1) \right] – \left[ – \frac{\mu_0 m m_p}{4\pi r^3} (2\cos\theta) \right] $$

Factor out the constant $C = \frac{\mu_0 m m_p}{4\pi r^3}$:

$$ W = C [ -(3\cos^2\theta – 1) + 2\cos\theta ] $$ $$ W = C [ 1 + 2\cos\theta – 3\cos^2\theta ] $$