Magnetic Dipole

It is known that the vector potential at $\vec{r}$ from the origin due to current $I$ is given by $$\vec{A}(\vec{r}) = \frac{\mu_0I}{4\pi} \oint\frac{1}{d}d\vec{l}$$ in which the $\frac{1}{d}$ can be expressed as $$\frac{1}{d} = \frac{1}{\sqrt{r^2 + (r')^2 - 2rr'\cos\theta}} = \frac{1}{r}\sum_{n=0}^{\infty}\left(\frac{r'}{r}\right)^nP_n(\cos\theta)$$ So, the vector potential can be expressed as $$\begin{align} \vec{A}(\vec{r}) &= \frac{\mu_0I}{4\pi}\oint \frac{1}{d}d\vec{l} \\ &= \frac{\mu_0I}{4\pi}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\oint(r')^nP_n(\cos\theta)d\vec{l} \\ &= \frac{\mu_0I}{4\pi}\left[\frac{1}{r}\oint d\vec{l} + \frac{1}{r^2}\oint r'\cos\theta d\vec{l} + \frac{1}{r^3}\oint(r')^2\left(\frac{3}{2}\cos^2\theta - \frac{1}{2}\right)d\vec{l} + ...\right] \end{align}$$ Note that the monopole term $$\frac{\mu_0I}{4\pi}\frac{1}{r}\oint d\vec{l}$$ is always zero since $$\oint d\vec{l} = 0$$ This is consistent with the fact that there exists no magnetic monopoles.
The dipole term $$\begin{align} \vec{A}_{dip}(\vec{r}) &= \frac{\mu_0I}{4\pi r^2}\oint r' \cos\theta d\vec{l} \\ &= \frac{\mu_0I}{4\pi r^2}\oint (\hat{\vec{r}}\cdot\vec{r}') d\vec{l} \\ &= -\frac{\mu_0I}{4\pi r^2}\left(\hat{\vec{r}}\times\int d\vec{a} \right) \\ \end{align}$$ Define the magnetic dipole $$\bbox[5px,border:2px solid #666] { \vec{m} = I\int d\vec{a} = I\vec{a}, }$$

where $\vec{a}$ is the vector area of the current loop enclosed.

When far away, the vector potential is approximately equal to the magnetic dipole potential.

When the current source is a perfect magnetic dipole, i.e. $m$ is finite while $a$ is zero, then the higher order terms in the multipole expansion are all zero, then the vector potential is exactly equal to the dipole potential.

The dipole term can be expressed as $$\bbox[5px,border:2px solid #666] { \vec{A}_{dip}(\vec{r}) = \frac{\mu_0I}{4\pi}\frac{\vec{m}\times\hat{\vec{r}}}{r^2} }$$ Let $\vec{A}$ be along the $\hat{\phi}$. The magnetic field by a dipole is $$\vec{B}_{\text{dip}}(\vec{r})=\nabla\times\vec{A}=\frac{\mu_0m}{4\pi r^3}(2\cos\theta\hat{r}+\sin\theta\hat{\theta})$$ But $$\begin{align} \vec{m}&=(\vec{m}\cdot\hat{r})\hat{r}+(\vec{m}\cdot\hat{\theta})\hat{\theta}\\ &=m\cos\theta\hat{r}-m\sin\theta\hat{\theta} \end{align}$$ The magentic field is then $$\begin{align} \vec{B}_{\text{dip}}(\vec{r})&=\frac{\mu_0m}{4\pi r^3}(2\cos\theta\hat{r}+\sin\theta\hat{\theta})\\ &=\frac{\mu_0}{4\pi r^3}(3m\cos\theta\hat{r}-(m\cos\theta\hat{r}-m\sin\theta\hat{\theta}))\\ \end{align}$$ So, $$\bbox[5px,border:2px solid #666] { \vec{B}_{\text{dip}}=\frac{\mu_0}{4\pi}\frac{1}{r^3}(3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}) }$$

Example (Griffiths Third Edition Ex 5.13)

Find the magnetic dipole moment of the "bookend-shaped" loop. All sides have length $w$ and it carries a current $I$.

Solution: Consider the wire as two square plane loops. Their wires on x-axis cancel each other. So the net magnetic dipole moment is $$\vec{m}=I\omega^2\hat{y}+I\omega^2\hat{z}$$

Checkpoint (Griffiths Third Edition Q5.35)

A phonograph record of radius $R$, carrying a uniform surface charge $\sigma$, is rotating at constant angular velocity $\omega$. Find its magnetic dipole moment.

$$\begin{align} dI &= \sigma v dr\\ &= \sigma \omega r dr\\ m &= \int \int I da\\ &= \int^R_0 \sigma \omega r (\pi r^2) dr\\ &= \pi\sigma\omega R^4/4 \end{align}$$

Torques on Magnetic Dipole

Current $I$ is flowing counterclockwise in a square loop in a magnetic field $\vec{B}$ with side $a$ perpendicular to the $\vec{B}$ and side $b$ parallel to $\vec{B}$. The magnetic force on the right $a$ side is $$F = IaB$$ pointing into the screen, while an equal magnitude magnetic force on the left $a$ is pointing out of the screen, so the net force is zero but the torque will be $$N = IaB\left(\frac{b}{2}\right) - IaB\left(\frac{-b}{2}\right) = IabB = mB$$ pointing upward. So, $$\bbox[5px,border:2px solid #666] { \vec{N} = \vec{m} \times \vec{B} }$$ If the loop is infinitesimal and the magnetic force is not uniform, the magnetic force on the dipole will be $$\bbox[5px,border:2px solid #666] { \vec{F} = \nabla(\vec{m}\cdot\vec{B}) }$$

Example (Griffiths Third Edition Q 6.1)

Calculate the torque exerted on the square loop due to the circular loop (assume $r$ is much larger than $a$ or $b$).

Solution: $$\vec{N} = \vec{m_2}\times\vec{B}_1$$ and $$\vec{B_1}=\frac{\mu_0}{4\pi}\frac{1}{r^3}(3(\vec{m_1}\cdot\vec{r})\hat{r}-\vec{m_1}),$$ where $\hat{r}=\hat{y}$, $\vec{m_1}=m_1\hat{z}=(\pi a^2I)\hat{z}$ and $\vec{m_2}=m_2\hat{y}=b^2I\hat{y}$. So, $$\vec{B_1}=-\frac{\mu_0}{4\pi}\frac{m_1}{r^3}\hat{z}$$ The torque is then $$\begin{align} \vec{N} &= -\frac{\mu_0}{4\pi}\frac{m_1m_2}{r^3}(\hat{y}\times\hat{z})\\ &= -\frac{\mu_0}{4\pi}\frac{m_1m_2}{r^3}\hat{x}\\ &= -\frac{\mu_0}{4}\frac{(abl)^2}{r^3}\hat{x} \end{align}$$

Checkpoint

Calculate the torque exerted on the circular loop due to the square loop (assume $r$ is much larger than $a$ or $b$).

$$\vec{N} = \vec{m_1}\times\vec{B}_2$$ and $$\vec{B_2}=\frac{\mu_0}{4\pi}\frac{1}{r^3}(3(\vec{m_2}\cdot\hat{r})\hat{r}-\vec{m_2}),$$ where $\hat{r}=-\hat{y}$, $\vec{m_1}=m_1\hat{z}=(\pi a^2I)\hat{z}$ and $\vec{m_2}=m_2\hat{y}=b^2I\hat{y}$. So, $$\vec{B_2}=\frac{\mu_0}{4\pi}\frac{m_2}{r^3}(-3-1)\hat{y}=\frac{\mu_0}{4\pi}\frac{m_2}{r^3}(-2)\hat{y}$$ The torque is then $$\begin{align} \vec{N} &= \frac{\mu_0}{4\pi}\frac{m_1m_2}{r^3}(-2)(\hat{z}\times\hat{y})\\ &= \frac{\mu_0}{4\pi}\frac{m_1m_2}{r^3}(2)\hat{x}\\ &= -\frac{\mu_0}{2}\frac{(abI)^2}{r^3}\hat{x} \end{align}$$

Magnetization

Magnetization is defined as the magnetic dipole moment per unit volume.