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.