Radiation from the magnetic dipoles of Earth and pulsars

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 23.

Because Earth’s magnetic north pole is not at the same place as the orbital north pole, a component of Earth’s magnetic dipole rotates as the planet rotates each day. This gives rise to radiation being emitted. Should we worry that Earth is losing a lot of energy in this way?

Suppose the angle (latitude difference) between true north and magnetic north is ${\psi}$ and the magnetic dipole moment is ${\mathbf{M}}$. Then the component of ${\mathbf{M}}$ that rotates has magnitude ${M\sin\psi}$ so we can write this component as

$\displaystyle \mathbf{M}_{rot}=M\sin\psi\left[\hat{\mathbf{x}}\cos\omega t+\hat{\mathbf{y}}\sin\omega t\right] \ \ \ \ \ (1)$

The power radiated by a time-varying magnetic dipole is

$\displaystyle P=\frac{\mu_{0}\ddot{m}^{2}}{6\pi c^{3}} \ \ \ \ \ (2)$

where ${m\left(t\right)}$ is the magnitude of the magnetic dipole moment. In our case

$\displaystyle \ddot{m}=\omega^{2}M\sin\psi\left(\cos^{2}\omega t+\sin^{2}\omega t\right)^{1/2}=\omega^{2}M\sin\psi \ \ \ \ \ (3)$

$\displaystyle P=\frac{\mu_{0}\omega^{4}M^{2}\sin^{2}\psi}{6\pi c^{3}} \ \ \ \ \ (4)$

The magnetic field due to a dipole moment ${\mathbf{M}}$ is

$\displaystyle \mathbf{B}=\frac{\mu_{0}}{4\pi r^{3}}\left[3\left(\mathbf{M}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{M}\right] \ \ \ \ \ (5)$

Taking Earth’s field strength at the equator to be around ${0.5\mbox{ Gauss}=5\times10^{-5}\mbox{ Tesla}}$ and taking ${\mathbf{M}\cdot\hat{\mathbf{r}}=0}$ at the equator and the radius of Earth to be ${r=6.37\times10^{6}\mbox{ m}}$, we get an estimate of Earth’s magnetic dipole moment:

$\displaystyle M\approx\frac{4\pi r^{3}B}{\mu_{0}}=1.3\times10^{23}\mbox{Amp m}^{2} \ \ \ \ \ (6)$

Using ${\psi=11^{\circ}}$ and ${\omega=2\pi/\left(60\times60\times24\right)=7.27\times10^{-5}\mbox{ s}^{-1}}$ we get

$\displaystyle P=4\times10^{-5}\mbox{ watts} \ \ \ \ \ (7)$

We needn’t worry about Earth losing a significant amount of energy through radiation from its magnetic field.

For a pulsar, however, it’s a different story. A typical pulsar has a radius of around ${10\mbox{ km}=10^{4}\mbox{ m}}$, a magnetic field strength of ${10^{8}\mbox{ Tesla}}$ and a rotation period of ${10^{-3}\mbox{ s}}$ giving ${\omega=2\pi\times10^{3}\mbox{ s}^{-1}}$. This gives

 $\displaystyle M$ $\displaystyle =$ $\displaystyle 10^{27}\mbox{ Amp m}^{2}\ \ \ \ \ (8)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle 3.8\times10^{36}\sin^{2}\psi\mbox{ watts} \ \ \ \ \ (9)$

Taking an average value of ${\frac{1}{2}}$ for ${\sin^{2}\psi}$ we get a power of around ${2\times10^{36}\mbox{ watts}}$. By comparison, the Sun produces ‘only’ around ${4\times10^{26}\mbox{ watts}}$ so a pulsar produces the output of 5 billion suns from its magnetic dipole radiation alone.

Not all pulsars generate this much power, though. The period of the crab nebula pulsar is around 33 ms, so (assuming the radius and magnetic field are the same as above) this gives a power of around ${2\times10^{30}\mbox{ watts}}$ which is still pretty impressive.

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 22.

A radio station’s transmitter consists of a circular current loop acting as a magnetic dipole. The dipole is on a radio tower 200 m above the ground, and is emitting a total power of ${3.5\times10^{4}\mbox{ watts}}$. We can find the position on the ground receiving the maximum power, although unlike the previous problem, we’re concerned only with the actual magnitude of the power and not with the power per unit area of ground. The formula for intensity is

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}m_{0}^{2}\omega^{4}\sin^{2}\theta}{32\pi^{2}c^{3}r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (1)$

For a location on the ground that is a distance ${R}$ from the base of the tower,

 $\displaystyle \sin\theta$ $\displaystyle =$ $\displaystyle \frac{R}{\sqrt{R^{2}+h^{2}}}\ \ \ \ \ (2)$ $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle R^{2}+h^{2} \ \ \ \ \ (3)$

so the intensity is

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}m_{0}^{2}\omega^{4}}{32\pi^{2}c^{3}}\frac{R^{2}}{\left(R^{2}+h^{2}\right)^{2}}\hat{\mathbf{r}} \ \ \ \ \ (4)$

The position on the ground receiving maximum intensity is determined from

 $\displaystyle \frac{d\left\langle S\right\rangle }{dR}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4R^{3}}{\left(R^{2}+h^{2}\right)^{3}}+\frac{2R}{\left(R^{2}+h^{2}\right)^{2}}\ \ \ \ \ (6)$ $\displaystyle R_{max}$ $\displaystyle =$ $\displaystyle h \ \ \ \ \ (7)$

and the intensity at this location is

 $\displaystyle \left\langle S\right\rangle _{max}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{128\pi^{2}c^{3}h^{2}} \ \ \ \ \ (8)$

The total power is obtained by integrating over a sphere of radius ${r}$:

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \int\mathbf{S}\cdot d\mathbf{a}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{32\pi^{2}c^{3}}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin^{2}\theta}{r^{2}}r^{2}\sin\theta d\phi d\theta\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \ \ \ \ \ (11)$

Therefore the maximum intensity can be written in terms of the power:

 $\displaystyle \left\langle S\right\rangle _{max}$ $\displaystyle =$ $\displaystyle \frac{12}{128\pi}\frac{\left\langle P\right\rangle }{h^{2}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{12}{128\pi}\frac{3.5\times10^{4}}{200^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.6\times10^{-2}\mbox{ watts m}^{-2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.6\times10^{-6}\mbox{ watts cm}^{-2} \ \ \ \ \ (15)$

Radiation from a current loop with time-varying current

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 12.

We’ve looked at the fields produced by a magnetic dipole that oscillates with a regular frequency ${\omega}$. By following the procedure in Griffiths’s section 11.1.4, where he derives the fields due to an electric dipole of arbitrary shape, we can derive the formulas for a magnetic dipole consisting of a circular current loop carrying a time-dependent current ${I\left(t\right)}$ where the time dependence is arbitrary.

We assume the current loop has radius ${b}$ and lies in the ${xy}$ plane with its centre on the ${z}$ axis. Since at any instant, the magnitude of the current is the same everywhere in the loop, we can use the same argument as in the oscillating case to deduce that for some observation point ${\mathbf{r}}$ in the ${xz}$ plane, the vector potential ${\mathbf{A}}$ points in the ${y}$ direction, and thus, since ${\mathbf{A}}$ is always tangential to the loop, its direction in general is in the ${\phi}$ direction. If the loop is electrically neutral, the electric potential is ${V=0}$, so we need to calculate only ${\mathbf{A}}$. The retarded potential is

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\frac{I\left(t-d/c\right)}{d}d\boldsymbol{\ell}'\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}b}{4\pi}\hat{\boldsymbol{\phi}}\int_{0}^{2\pi}\frac{I\left(t-d/c\right)}{d}\cos\phi'd\phi' \ \ \ \ \ (2)$

where ${\phi'}$ is the azimuthal angle around the loop so that the ${y}$ component of ${d\boldsymbol{\ell}'}$ is ${b\cos\phi'}$ and the retarded time is

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (3)$

and

 $\displaystyle d$ $\displaystyle \equiv$ $\displaystyle \left|\mathbf{r}-\mathbf{r}'\right|\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}}\ \ \ \ \ (5)$ $\displaystyle \hat{\mathbf{d}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{r}-\mathbf{r}'}{d} \ \ \ \ \ (6)$

where ${\mathbf{r}'}$ is the position on the loop being integrated over.

For our observation point in the ${xz}$ plane, we have

$\displaystyle \mathbf{r}=r\sin\theta\hat{\mathbf{x}}+r\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (7)$

and for a point on the loop

$\displaystyle \mathbf{r}'=b\cos\phi'\hat{\mathbf{x}}+b\sin\phi'\hat{\mathbf{y}} \ \ \ \ \ (8)$

In what follows, we’ll use the notation ${c_{\theta}\equiv\cos\theta}$, ${s_{\theta}\equiv\sin\theta}$, etc to simplify the notation.

Therefore, assuming ${b\ll r}$ (the loop is very small)

 $\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}+b^{2}-2\mathbf{r}\cdot\mathbf{r}'}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle r\left(1-\frac{b}{r}s_{\theta}c_{\phi'}\right)\ \ \ \ \ (10)$ $\displaystyle \frac{1}{d}$ $\displaystyle \cong$ $\displaystyle \frac{1}{r}\left(1+\frac{b}{r}s_{\theta}c_{\phi'}\right) \ \ \ \ \ (11)$

We can expand the current in a Taylor series about ${t_{0}\equiv t-\frac{r}{c}}$:

 $\displaystyle I\left(t-\frac{d}{c}\right)$ $\displaystyle \cong$ $\displaystyle I\left(t-\frac{r}{c}+\frac{b}{c}s_{\theta}c_{\phi'}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I\left(t_{0}\right)+\dot{I}\left(t_{0}\right)\frac{b}{c}s_{\theta}c_{\phi'}+\frac{1}{2!}\ddot{I}\left(t_{0}\right)\left(\frac{b}{c}s_{\theta}c_{\phi'}\right)^{2}+\dots \ \ \ \ \ (13)$

We are justified in dropping the last term if

 $\displaystyle \frac{1}{2!}\ddot{I}\left(t_{0}\right)\left(\frac{b}{c}s_{\theta}c_{\phi'}\right)^{2}$ $\displaystyle \ll$ $\displaystyle \dot{I}\left(t_{0}\right)\frac{b}{c}s_{\theta}c_{\phi'}\ \ \ \ \ (14)$ $\displaystyle b$ $\displaystyle \ll$ $\displaystyle \frac{c}{\left|\ddot{I}/\dot{I}\right|} \ \ \ \ \ (15)$

If we compare higher derivative terms with the first order term, we get the general condition

$\displaystyle b\ll c\left|\frac{\dot{I}}{d^{n}I/dt^{n}}\right|^{n-1} \ \ \ \ \ (16)$

Assuming this is true, we get from 2:

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}b}{4\pi r}\hat{\boldsymbol{\phi}}\int_{0}^{2\pi}\left(I\left(t_{0}\right)+\dot{I}\left(t_{0}\right)\frac{b}{c}s_{\theta}c_{\phi'}\right)\left(1+\frac{b}{r}s_{\theta}c_{\phi'}\right)c_{\phi'}d\phi'\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle \frac{\mu_{0}\pi b^{2}}{4\pi r}\left(\frac{I\left(t_{0}\right)}{r}+\frac{\dot{I}\left(t_{0}\right)}{c}\right)s_{\theta}\hat{\boldsymbol{\phi}} \ \ \ \ \ (18)$

where to get the second line, we discarded the term in ${b^{3}}$ and used ${\int_{0}^{2\pi}\cos\phi'd\phi'=0}$ and ${\int_{0}^{2\pi}\cos^{2}\phi'd\phi'=\pi}$. If we’re interested only in the radiation produced by this dipole, we can ignore any terms in the potential that are of order 2 or higher in ${\frac{1}{r}}$, since it is only ${\frac{1}{r^{2}}}$ terms in the Poynting vector that will contribute to radiation that escapes to infinity. Therefore, we can throw away the first term above to get our final approximation:

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)\cong\frac{\mu_{0}\pi b^{2}}{4\pi rc}\dot{I}\left(t_{0}\right)s_{\theta}\hat{\boldsymbol{\phi}} \ \ \ \ \ (19)$

We can write this in terms of the magnetic moment of the loop, which is

$\displaystyle m\left(t_{0}\right)=\pi b^{2}I\left(t_{0}\right) \ \ \ \ \ (20)$

so

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)\cong\frac{\mu_{0}}{4\pi rc}\dot{m}\left(t_{0}\right)s_{\theta}\hat{\boldsymbol{\phi}}$

We can now calculate the fields:

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}s_{\theta}}{4\pi rc}\ddot{m}\left(t_{0}\right)\hat{\boldsymbol{\phi}}\ \ \ \ \ (22)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{r}\frac{\partial}{\partial r}\left(rA\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}s_{\theta}}{4\pi rc}\frac{\partial\dot{m}}{\partial r}\hat{\boldsymbol{\theta}} \ \ \ \ \ (25)$

where in calculating ${\mathbf{B}}$, we ignored the term ${\frac{1}{rs_{\theta}}\frac{\partial}{\partial\theta}\left(s_{\theta}A\right)\hat{\mathbf{r}}}$ since it gives a term containing ${\frac{1}{r^{2}}}$.

Since ${m=m\left(t-\frac{r}{c}\right)}$ we have

 $\displaystyle \frac{\partial\dot{m}}{\partial r}$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\ddot{m}\ \ \ \ \ (26)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}s_{\theta}}{4\pi rc^{2}}\ddot{m}\hat{\boldsymbol{\theta}} \ \ \ \ \ (27)$

The Poynting vector is

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}s_{\theta}^{2}\ddot{m}^{2}}{16\pi^{2}r^{2}c^{3}}\hat{\mathbf{r}} \ \ \ \ \ (29)$

The power radiated is the integral of ${\mathbf{S}}$ over a large sphere of radius ${r}$:

 $\displaystyle P$ $\displaystyle =$ $\displaystyle \int\mathbf{S}\cdot d\mathbf{a}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi\mu_{0}\ddot{m}^{2}}{16\pi^{2}c^{3}}\int_{0}^{\pi}\frac{\sin^{2}\theta}{r^{2}}r^{2}\sin\theta d\theta\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\ddot{m}^{2}}{6\pi c^{3}} \ \ \ \ \ (32)$

Radiation from a magnetic dipole composed of monopoles

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 7

We’ve seen that the fields produced by an oscillating magnetic dipole are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left\{ \frac{\omega^{2}}{c}\cos\left[\omega\left(t-r/c\right)\right]+\frac{\omega}{r}\sin\left[\omega\left(t-r/c\right)\right]\right\} \hat{\boldsymbol{\phi}}\ \ \ \ \ (2)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\cos\theta}{2\pi r^{2}}\left[\frac{1}{r}\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{c}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\mathbf{r}}+\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left[\left(\frac{1}{r^{2}}-\frac{\omega^{2}}{c^{2}}\right)\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{rc}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\theta}}\nonumber$

By making the approximation that the observation distance ${r}$ is much larger than the wavelength of radiation, so that ${r\gg c/\omega}$, these formulas simplify to

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi cr}\hat{\boldsymbol{\phi}}\cos\left[\omega\left(t-r/c\right)\right]\ \ \ \ \ (5)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi c^{2}r}\hat{\boldsymbol{\theta}}\cos\left[\omega\left(t-r/c\right)\right] \ \ \ \ \ (6)$

Now let’s return to the fantasy world where magnetic monopoles exist, so that another way we can create a magnetic dipole is to connect two magnetic charges by a wire and then drive charge back and forth between the ends of the wire, in the same way that we did for the electric dipole. Earlier, we’ve seen that if we include magnetic charge in Maxwell’s equations, the duality transformation produces fields that still satisfy Maxwell’s equations:

 $\displaystyle \mathbf{E}^{\prime}$ $\displaystyle =$ $\displaystyle \mathbf{E}\cos\alpha+c\mathbf{B}\sin\alpha\ \ \ \ \ (7)$ $\displaystyle c\mathbf{B}^{\prime}$ $\displaystyle =$ $\displaystyle c\mathbf{B}\cos\alpha-\mathbf{E}\sin\alpha\ \ \ \ \ (8)$ $\displaystyle cq_{e}^{\prime}$ $\displaystyle =$ $\displaystyle cq_{e}\cos\alpha+q_{m}\sin\alpha\ \ \ \ \ (9)$ $\displaystyle q_{m}^{\prime}$ $\displaystyle =$ $\displaystyle q_{m}\cos\alpha-cq_{e}\sin\alpha \ \ \ \ \ (10)$

where ${\alpha}$ is a rotation angle in ${\mathbf{E}-\mathbf{B}}$ space. If we start with the fields generated by an oscillating electric dipole and then choose ${\alpha=\frac{\pi}{2}}$ so that we convert all electric charge into magnetic charge, we can generate the fields that would be produced by an oscillating magnetic dipole constructed using magnetic charge as described above. The original fields for the electric dipole are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (12)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi c}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\phi}} \ \ \ \ \ (14)$

The required transformations with ${\alpha=\frac{\pi}{2}}$ are

 $\displaystyle \mathbf{E}'$ $\displaystyle =$ $\displaystyle c\mathbf{B}\ \ \ \ \ (15)$ $\displaystyle c\mathbf{B}'$ $\displaystyle =$ $\displaystyle -\mathbf{E}\ \ \ \ \ (16)$ $\displaystyle cq_{e}'$ $\displaystyle =$ $\displaystyle q_{m}\ \ \ \ \ (17)$ $\displaystyle q_{m}'$ $\displaystyle =$ $\displaystyle -cq_{e} \ \ \ \ \ (18)$

As the electric dipole moment ${p_{0}=q_{e}l}$ is the product of an electric charge ${q_{e}}$ and the length ${l}$ of the wire joining the two charges, it transforms in the same way as ${q_{e}}$ so we have, if we take the magnetic moment to be ${m_{0}=q_{m}'l}$:

$\displaystyle m_{0}=-cp_{0} \ \ \ \ \ (19)$

Applying these transformations to 12 and 14 we get

 $\displaystyle \mathbf{E}'$ $\displaystyle =$ $\displaystyle -c\frac{\mu_{0}\left(-m_{0}/c\right)\omega^{2}}{4\pi c}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\phi}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi cr}\hat{\boldsymbol{\phi}}\cos\left[\omega\left(t-r/c\right)\right]\ \ \ \ \ (21)$ $\displaystyle \mathbf{B}'$ $\displaystyle =$ $\displaystyle -\left(-\frac{1}{c}\right)\frac{\mu_{0}\left(-m_{0}/c\right)\omega^{2}}{4\pi}\frac{\sin\theta}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\hat{\boldsymbol{\theta}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}m_{0}\omega^{2}\sin\theta}{4\pi c^{2}r}\hat{\boldsymbol{\theta}}\cos\left[\omega\left(t-r/c\right)\right] \ \ \ \ \ (23)$

These fields are the same as 5 and 6 that we got from the current loop. Thus we can’t tell whether magnetic dipole radiation is coming from a current loop or from magnetic monopoles.

Radiation resistance of an oscillating magnetic dipole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 6

We can find the radiation resistance of an oscillating magnetic dipole produced by an AC current in a circular wire loop of radius ${b}$ in the same way as for an oscillating electric dipole. The average power radiated by the magnetic dipole is the integral of the intensity over a sphere, so we have

 $\displaystyle \left\langle P\right\rangle$ $\displaystyle =$ $\displaystyle \int\left\langle \mathbf{S}\right\rangle \cdot d\mathbf{a}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\frac{\mu_{0}m_{0}^{2}\omega^{4}\sin^{2}\theta}{32\pi^{2}c^{3}r^{2}}\hat{\mathbf{r}}\cdot d\mathbf{a}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{32\pi^{2}c^{3}}2\pi\int_{0}^{\pi}\frac{\sin^{2}\theta}{r^{2}}r^{2}\sin\theta d\theta\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \ \ \ \ \ (4)$

The resistance ${R}$ required in the wire loop to generate the same power loss through heat is given by ${P=\left\langle I^{2}\right\rangle R}$ where the current is

$\displaystyle I\left(t\right)=I_{0}\cos\omega t \ \ \ \ \ (5)$

with the maximum current ${I_{0}}$ given in terms of the maximum magnetic moment ${m_{0}}$:

$\displaystyle I_{0}=\frac{m_{0}}{\pi b^{2}} \ \ \ \ \ (6)$

Since the average of ${\cos^{2}x}$ over a complete cycle is ${\frac{1}{2}}$ we get

 $\displaystyle \left\langle I^{2}\right\rangle R$ $\displaystyle =$ $\displaystyle \frac{m_{0}^{2}}{2\pi^{2}b^{4}}R\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}}\ \ \ \ \ (8)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}\pi\omega^{4}b^{4}}{6c^{3}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\mu_{0}\pi^{5}c}{3}\frac{b^{4}}{\lambda^{4}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.08\times10^{5}\frac{b^{4}}{\lambda^{4}}\;\Omega \ \ \ \ \ (11)$

where in the fourth line we used ${\omega/c=2\pi/\lambda}$.

To compare this to the electric dipole’s radiation resistance, which is given in terms of the length ${l}$ of the wire joining the two charges as:

$\displaystyle R_{e}=787\frac{l^{2}}{\lambda^{2}}\;\Omega \ \ \ \ \ (12)$

we can take ${l=2\pi b}$ so the lengths of wire in the two cases are the same. Then

 $\displaystyle \frac{R_{e}}{R_{m}}$ $\displaystyle =$ $\displaystyle \frac{787\left(2\pi\right)^{2}}{3.08\times10^{5}}\frac{\lambda^{2}}{b^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.1\frac{\lambda^{2}}{b^{2}} \ \ \ \ \ (14)$

Since we’re assuming that ${b\ll\lambda}$, ${R_{e}\gg R_{m}}$.

Fields of an oscillating magnetic dipole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 5

We can analyze an oscillating magnetic dipole in a similar way to the electric dipole. We begin with a small circular current loop of radius ${b}$ in the ${xy}$ plane, centred at the origin. The current is driven to be alternating, so that

$\displaystyle I\left(t\right)=I_{0}\cos\omega t \ \ \ \ \ (1)$

The magnetic dipole moment of a current loop is

$\displaystyle \mathbf{m}=I\mathbf{a} \ \ \ \ \ (2)$

where ${\mathbf{a}}$ is the vector area of the loop, which for a planar circular loop is just ${\pi b^{2}\hat{\mathbf{z}}}$. Thus

 $\displaystyle \mathbf{m}\left(t\right)$ $\displaystyle =$ $\displaystyle \pi b^{2}I_{0}\cos\left(\omega t\right)\hat{\mathbf{z}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle m_{0}\cos\left(\omega t\right)\hat{\mathbf{z}} \ \ \ \ \ (4)$

If the loop is electrically neutral, the electric potential is ${V=0}$, so we need to calculate only ${\mathbf{A}}$. The retarded potential is

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)=\frac{\mu_{0}}{4\pi}\int\frac{I_{0}\cos\omega\left(t-d/c\right)}{d}d\boldsymbol{\ell}' \ \ \ \ \ (5)$

where the integral is taken around the loop and the retarded time is

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (6)$

and

 $\displaystyle d$ $\displaystyle \equiv$ $\displaystyle \left|\mathbf{r}-\mathbf{r}'\right|\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}}\ \ \ \ \ (8)$ $\displaystyle \hat{\mathbf{d}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{r}-\mathbf{r}'}{d} \ \ \ \ \ (9)$

where ${\mathbf{r}'}$ is the position on the loop being integrated over.

To work out the integral, we can start by considering ${\mathbf{r}}$ to be some point in the ${xz}$ plane. The line integral can be broken down into pairs of increments on the circle located at ${\left(x,\pm y,0\right)}$, that is, each pair of points is symmetric about the ${x}$ axis. The increment ${d\boldsymbol{\ell}_{-}}$ at ${\left(x,-y,0\right)}$ has components ${\left(dx,dy,0\right)}$ while the increment ${d\boldsymbol{\ell}_{+}}$ at ${\left(x,y,0\right)}$ has components ${\left(-dx,dy,0\right)}$. Thus the ${x}$ components cancel in pairs in the integral, while the ${y}$ components add in pairs, so the net result of the integral around the entire circle is a vector pointing in the ${y}$ direction. Since the circle is symmetric about the ${z}$ axis, we can generalize this result to deduce that ${\mathbf{A}}$ has a direction that is always tangential to the circle, which means that, in spherical coordinates, it is in the ${\phi}$ direction and has the same magnitude at all points around the circle.

Griffiths goes through the calculation of ${\mathbf{A}}$ in detail in his section 11.1.3 so I won’t repeat that here, other than to note that he uses the same approximations as were used with the electric dipole, namely that the radius of the loop ${b}$ is much less than the observation distance ${r}$, and that ${b}$ is also much smaller than the wavelength of the radiation, represented by the condition ${b\ll c/\omega}$. The result is

$\displaystyle \mathbf{A}\left(r,\theta,t\right)=\frac{\mu_{0}m_{0}}{4\pi}\frac{\sin\theta}{r}\left[\frac{1}{r}\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{c}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\phi}} \ \ \ \ \ (10)$

At this stage, Griffiths invokes a further approximation by assuming that ${r\gg c/\omega}$ (observer is much further away than the wavelength of the radiation). However, we can calculate the fields without making that approximation to see how much of an effect that approximation has. Since ${V=0}$ we have

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left\{ \frac{\omega^{2}}{c}\cos\left[\omega\left(t-r/c\right)\right]+\frac{\omega}{r}\sin\left[\omega\left(t-r/c\right)\right]\right\} \hat{\boldsymbol{\phi}}\ \ \ \ \ (12)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}m_{0}\cos\theta}{2\pi r^{2}}\left[\frac{1}{r}\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{c}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\mathbf{r}}+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}m_{0}\sin\theta}{4\pi r}\left[\left(\frac{1}{r^{2}}-\frac{\omega^{2}}{c^{2}}\right)\cos\left[\omega\left(t-r/c\right)\right]-\frac{\omega}{rc}\sin\left[\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\theta}}\nonumber$

These fields have the same form as those we worked out earlier for a spherical wave in vacuum.

The Poynting vector can be worked out and simplified using Maple to combine the trig products using double angle formulas:

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}m_{0}^{2}\sin^{2}\theta}{32\pi^{2}c^{3}r^{5}}\left[\left(2c^{2}\omega^{2}r-\omega^{4}r^{3}\right)\cos\left[2\omega\left(t-r/c\right)\right]+\left(\omega c^{3}-2\omega^{3}cr^{2}\right)\sin\left[2\omega\left(t-r/c\right)\right]-\omega^{4}r^{3}\right]\hat{\mathbf{r}}+\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\mu_{0}m_{0}^{2}\sin\left(2\theta\right)}{32\pi^{2}c^{3}r^{5}}\left[2c^{2}\omega^{2}r\cos\left[2\omega\left(t-r/c\right)\right]+\left(\omega c^{3}-\omega^{3}cr^{2}\right)\sin\left[2\omega\left(t-r/c\right)\right]\right]\hat{\boldsymbol{\theta}}\nonumber$

To get the average intensity, we integrate ${\mathbf{S}}$ over a complete cycle, that is, for ${t=0}$ to ${t=2\pi/\omega}$ and then multiply by ${\omega/2\pi}$ to get the average. Integrating over one cycle causes each of the double angle trig functions to go through two complete cycles, so they all integrate to zero and the only term that is left is the ${-\omega^{4}r^{3}}$ term in the radial component, so we get

$\displaystyle \left\langle \mathbf{S}\right\rangle =\frac{\mu_{0}m_{0}^{2}\omega^{4}\sin^{2}\theta}{32\pi^{2}c^{3}r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (17)$

This value is the same as that obtained by assuming ${r\gg c/\omega}$ from the start (equation 11.39 in Griffiths).

Mutual inductance between two magnetic dipoles

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 7.30.

This problem relates mutual inductance to the interaction energy of two magnetic dipoles. Suppose we have two tiny loops of wire with area vectors ${\mathbf{a}_{1}}$ and ${\mathbf{a}_{2}}$. The loops become dipoles if they carry a current ${I}$ so that their magnetic dipole moments are ${\mathbf{m}_{i}=I\mathbf{a}_{i}}$ (${i=1,2}$). We can work out the mutual inductance between these two loops if we assume that the areas are small enough that the field produced by one dipole is constant over the area of the other dipole. The magnetic field of dipole 2 as felt by dipole 1 is

 $\displaystyle \mathbf{B}_{21}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r_{12}^{3}}\left[3\left(\mathbf{m}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\hat{\mathbf{r}}_{21}-\mathbf{m}_{2}\right]\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{4\pi r_{12}^{3}}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\hat{\mathbf{r}}_{21}-\mathbf{a}_{2}\right] \ \ \ \ \ (2)$

where ${\hat{\mathbf{r}}_{21}}$ is the unit vector pointing from dipole 2 to dipole 1. The mutual inductance can be worked out if we know the flux through dipole 1. With our approximation of constant field over dipole 1, we get

 $\displaystyle \Phi_{21}$ $\displaystyle =$ $\displaystyle \mathbf{B}_{21}\cdot\mathbf{a}_{1}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{4\pi r_{12}^{3}}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{a}_{1}\right)-\mathbf{a}_{2}\cdot\mathbf{a}_{1}\right] \ \ \ \ \ (4)$

The mutual inductance is

$\displaystyle M_{21}=\frac{\Phi_{21}}{I}=\frac{\mu_{0}}{4\pi r_{12}^{3}}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{a}_{1}\right)-\mathbf{a}_{2}\cdot\mathbf{a}_{1}\right] \ \ \ \ \ (5)$

Conversely, if we run a current ${I}$ through dipole 2 and look at the flux through dipole 1, we get

 $\displaystyle \Phi_{12}$ $\displaystyle =$ $\displaystyle \mathbf{B}_{12}\cdot\mathbf{a}_{2}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I}{4\pi r_{12}^{3}}\left[3\left(\mathbf{a}_{1}\cdot\hat{\mathbf{r}}_{12}\right)\left(\hat{\mathbf{r}}_{12}\cdot\mathbf{a}_{2}\right)-\mathbf{a}_{1}\cdot\mathbf{a}_{2}\right] \ \ \ \ \ (7)$

Since ${\hat{\mathbf{r}}_{12}=-\hat{\mathbf{r}}_{21}}$, we see that ${\Phi_{12}=\Phi_{21}}$ and thus ${M_{12}=M_{21}}$ as required.

If we now start off with a current ${I_{1}}$ flowing in dipole 1 but no current in dipole 2, then switch on a current ${I_{2}}$ in dipole 2, the emf induced in dipole 1 is given by the change in flux:

 $\displaystyle \mathcal{E}_{21}$ $\displaystyle =$ $\displaystyle -\frac{d\Phi_{21}}{dt}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -M_{21}\frac{dI_{2}}{dt}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{4\pi r_{12}^{3}}\frac{dI_{2}}{dt}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{a}_{1}\right)-\mathbf{a}_{2}\cdot\mathbf{a}_{1}\right] \ \ \ \ \ (10)$

To get the total work required if we want to maintain ${I_{1}}$ in dipole 1, we need to consider how this induced emf affects ${I_{1}}$. If we didn’t attempt to maintain ${I_{1}}$, then ${I_{1}}$ would change in the direction prescribed by the sign of ${\mathcal{E}_{21}}$. So if we want to maintain ${I_{1}}$, we have to oppose ${\mathcal{E}_{21}}$, and thus the rate at which the work is done is ${dW_{21}/dt=-\mathcal{E}_{21}I_{1}}$, where the minus sign indicates we’re opposing ${\mathcal{E}_{21}}$.

 $\displaystyle \frac{dW_{21}}{dt}$ $\displaystyle =$ $\displaystyle -\mathcal{E}_{21}I_{1}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I_{1}}{4\pi r_{12}^{3}}\frac{dI_{2}}{dt}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{a}_{1}\right)-\mathbf{a}_{2}\cdot\mathbf{a}_{1}\right] \ \ \ \ \ (12)$

Since we’re keeping ${I_{1}}$ and all the geometrical terms constant, we can just integrate this expression to get the total work done:

 $\displaystyle W_{21}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I_{1}}{4\pi r_{12}^{3}}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{a}_{1}\right)-\mathbf{a}_{2}\cdot\mathbf{a}_{1}\right]\int_{0}^{t}\frac{dI_{2}}{dt}dt\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}I_{1}I_{2}}{4\pi r_{12}^{3}}\left[3\left(\mathbf{a}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{a}_{1}\right)-\mathbf{a}_{2}\cdot\mathbf{a}_{1}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r_{12}^{3}}\left[3\left(\mathbf{m}_{2}\cdot\hat{\mathbf{r}}_{21}\right)\left(\hat{\mathbf{r}}_{21}\cdot\mathbf{m}_{1}\right)-\mathbf{m}_{2}\cdot\mathbf{m}_{1}\right] \ \ \ \ \ (15)$

This last expression is exactly equal to the interaction energy between the two dipoles except that it’s opposite in sign. I’m not really sure why this should be (assuming this answer is correct), though the two cases are physically different. In this case, we start with one dipole at full strength given by ${\mathbf{m}_{1}}$ and then turn the other one on, so the work doesn’t involve any movement of the dipoles; we can move dipole 2 into position before we turn the current on, so there is no interaction with the first dipole until it’s in place. In the previous case, both dipoles were switched on and then moved into position, so the motion of the dipoles causes the change in flux (and thus the back emf), rather than any change in the current (and hence in the dipoles’ strengths). Not sure.

Energy of magnetic dipole in magnetic field

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.21.

Using similar techniques to those for finding the energy of an electric dipole in an electric field, we can work out the energy of a magnetic dipole in a magnetic field. The torque exerted by a field on a dipole is

$\displaystyle \mathbf{N}=\mathbf{m}\times\mathbf{B} \ \ \ \ \ (1)$

If we bring in a dipole from infinity in such a way that no work is done (by moving the dipole along a path that is always perpendicular to the force exerted by the field) then when we arrive at the desired location, we must rotate the dipole into its final position, and this requires work to be done against the torque. If we start with ${\mathbf{m}\perp\mathbf{B}}$, then the work done is

$\displaystyle U=mB\int_{\pi/2}^{\theta}\sin\theta d\theta=-mB\cos\theta=-\mathbf{m}\cdot\mathbf{B} \ \ \ \ \ (2)$

The choice of ${\theta=\pi/2}$ as the zero point for the energy appears to be arbitrary, although it does make the situation more symmetric. The maximum energy of ${mB}$ occurs when ${\mathbf{m}}$ is anti-parallel to ${\mathbf{B}}$ and the minimum energy of ${-mB}$ when ${\mathbf{m}}$ and ${\mathbf{B}}$ are parallel. If we had chosen a different zero angle, the difference in energy between two angles would still be the same, and that is all that matters physically.

The interaction energy of two dipoles can be written down from the formula for the field due to a dipole:

$\displaystyle \mathbf{B}_{1}=\frac{\mu_{0}}{4\pi r^{3}}\left[3\left(\mathbf{m}_{1}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{m}_{1}\right] \ \ \ \ \ (3)$

We then get

 $\displaystyle U_{12}$ $\displaystyle =$ $\displaystyle -\mathbf{m}_{2}\cdot\mathbf{B}_{1}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r^{3}}\left[\mathbf{m}_{1}\cdot\mathbf{m}_{2}-3\left(\mathbf{m}_{1}\cdot\hat{\mathbf{r}}\right)\left(\mathbf{m}_{2}\cdot\hat{\mathbf{r}}\right)\right] \ \ \ \ \ (5)$

Here ${\hat{\mathbf{r}}}$ is the unit vector pointing from ${\mathbf{m}_{1}}$ to ${\mathbf{m}_{2}}$, so if we define ${\theta_{i}}$ to be the angle between ${\hat{\mathbf{r}}}$ and ${\mathbf{m}_{i}}$ then

$\displaystyle U_{12}=\frac{\mu_{0}m_{1}m_{2}}{4\pi r^{3}}\left(\cos\left(\theta_{1}-\theta_{2}\right)-3\cos\theta_{1}\cos\theta_{2}\right) \ \ \ \ \ (6)$

The stable configuration is when this energy is a minimum, so considering the trig functions alone, we have

 $\displaystyle \frac{\partial U_{12}}{\partial\theta_{1}}$ $\displaystyle =$ $\displaystyle -\sin\left(\theta_{1}-\theta_{2}\right)+3\sin\theta_{1}\cos\theta_{2}=0\ \ \ \ \ (7)$ $\displaystyle \frac{\partial U_{12}}{\partial\theta_{2}}$ $\displaystyle =$ $\displaystyle \sin\left(\theta_{1}-\theta_{2}\right)+3\cos\theta_{1}\sin\theta_{2}=0 \ \ \ \ \ (8)$

Using the shorthand notation ${s_{12}\equiv\sin\left(\theta_{1}-\theta_{2}\right)}$, ${s_{1}\equiv\sin\theta_{1}}$, ${c_{1}\equiv\cos\theta_{1}}$, etc, we have

 $\displaystyle s_{12}$ $\displaystyle =$ $\displaystyle 3s_{1}c_{2}\ \ \ \ \ (9)$ $\displaystyle s_{1}c_{2}+c_{1}s_{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (10)$ $\displaystyle \tan\theta_{1}$ $\displaystyle =$ $\displaystyle -\tan\theta_{2} \ \ \ \ \ (11)$

Since ${0\le\theta_{1},\theta_{2}\le\pi}$, this means that either ${\tan\theta_{1}=\tan\theta_{2}=0}$ or ${\theta_{1}=\frac{\pi}{2}+x;\;\theta_{2}=\frac{\pi}{2}-x}$. Either way we have ${s_{1}=s_{2}\equiv s}$ and ${c_{1}=-c_{2}\equiv c}$, so

 $\displaystyle s_{12}$ $\displaystyle =$ $\displaystyle -3sc\ \ \ \ \ (12)$ $\displaystyle s_{1}c_{2}-c_{1}s_{2}$ $\displaystyle =$ $\displaystyle -3sc\ \ \ \ \ (13)$ $\displaystyle -sc-cs$ $\displaystyle =$ $\displaystyle -3sc\ \ \ \ \ (14)$ $\displaystyle -2sc$ $\displaystyle =$ $\displaystyle -3sc \ \ \ \ \ (15)$

The only way the last equation can be true is if ${s=0}$ and/or ${c=0}$, which means the choices are ${\theta_{1,2}=0,\frac{\pi}{2},\pi}$. We can try the various possibilities, but it’s fairly obvious anyway that the stable configuration is when the two moments are parallel, that is, ${\theta_{1}=\theta_{2}=0}$ (or ${\theta_{1}=\theta_{2}=\pi}$ ). If we line up a sequence of compass needles, they will align themselves so that the north end of one points to the south end of the next, and so forth.

Force on a magnetic dipole – a better derivation

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.22.

We’ve seen a rather crude derivation of the force on a magnetic dipole in a varying magnetic field, which gives the result

$\displaystyle \mathbf{F}=\nabla\left(\mathbf{m}\cdot\mathbf{B}\right) \ \ \ \ \ (1)$

In that derivation, we considered a particular geometry of dipole (a square) and constrained the orientation of the dipole within the coordinate system. A more general approach is given here.

If ${\mathbf{B}\left(\mathbf{r}\right)}$ is a general vector field, we can write its value near the location ${\mathbf{r}_{0}}$ of the dipole as a 3-d Taylor expansion to first order:

$\displaystyle \mathbf{B}\left(\mathbf{r}\right)=\mathbf{B}\left(\mathbf{r}_{0}\right)+\left[\left(\mathbf{r}-\mathbf{r}_{0}\right)\cdot\nabla\right]\mathbf{B}_{0} \ \ \ \ \ (2)$

where the subscript 0 on ${\mathbf{B}_{0}}$ in the second term indicates that the derivatives of ${\mathbf{B}}$ are evaluated at ${\mathbf{r}_{0}}$.

Putting this into the Lorentz force law, we have (given the current ${I}$ producing the dipole)

 $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle I\oint d\boldsymbol{\ell}\times\mathbf{B}\left(\mathbf{r}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I\oint d\boldsymbol{\ell}\times\mathbf{B}\left(\mathbf{r}_{0}\right)+I\oint d\boldsymbol{\ell}\times\left[\left(\mathbf{r}-\mathbf{r}_{0}\right)\cdot\nabla\right]\mathbf{B}_{0} \ \ \ \ \ (4)$

The first integral and ${\mathbf{r}_{0}}$ term in the second integral come out to zero, since we are integrating a constant around a closed loop. Thus we are left with

$\displaystyle \mathbf{F}=I\oint d\boldsymbol{\ell}\times\left(\mathbf{r}\cdot\nabla\right)\mathbf{B}_{0} \ \ \ \ \ (5)$

We can write the cross product in rectangular coordinates using the Levi-Civita symbol ${\epsilon_{ijk}=+1}$ for a cyclic permutation of 1,2,3, ${-1}$ for an anti-cyclic permutation and zero if any two indices are equal. In general

$\displaystyle \left(\mathbf{A}\times\mathbf{B}\right)_{i}=\sum_{j,k=1}^{3}\epsilon_{ijk}A_{j}B_{k} \ \ \ \ \ (6)$

The dot product can be written as

$\displaystyle \mathbf{A}\cdot\mathbf{B}=\sum_{l=1}^{3}A_{l}B_{l} \ \ \ \ \ (7)$

Using these two sums, we have for a component of the force:

$\displaystyle F_{i}=I\sum_{j,k,l}\epsilon_{ijk}\oint r_{l}d\ell_{j}\nabla_{l}B_{k0} \ \ \ \ \ (8)$

The integral is of a form that we’ve encountered when discussing the vector area ${\mathbf{a}}$ of a curve. For a constant vector ${\mathbf{c}}$:

$\displaystyle \oint\mathbf{c}\cdot\mathbf{r}d\boldsymbol{\ell}=\mathbf{a}\times\mathbf{c} \ \ \ \ \ (9)$

In components, this is

$\displaystyle \sum_{l}\oint r_{l}c_{l}d\ell_{j}=\sum_{m,n}\epsilon_{jmn}a_{m}c_{n} \ \ \ \ \ (10)$

Using ${\mathbf{c}=\nabla B_{k0}}$, we can plug this back into the force equation to get

$\displaystyle F_{i}=I\sum_{j,k,m,n}\epsilon_{ijk}\epsilon_{jmn}a_{m}\left(\nabla B_{k0}\right)_{n} \ \ \ \ \ (11)$

We can now use an identity for a sum over the Levi-Civita symbols:

$\displaystyle \sum_{j}\epsilon_{ijk}\epsilon_{njm}=\delta_{in}\delta_{km}-\delta_{im}\delta_{kn} \ \ \ \ \ (12)$

Because of the cyclic property of the ${\epsilon_{njm}}$, we have ${\epsilon_{njm}=\epsilon_{jmn}}$, so we now get for the force:

 $\displaystyle F_{i}$ $\displaystyle =$ $\displaystyle I\sum_{k,m,n}\left(\delta_{in}\delta_{km}-\delta_{im}\delta_{kn}\right)a_{m}\left(\nabla B_{k0}\right)_{n}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I\sum_{k}a_{k}\left(\nabla B_{k0}\right)_{i}-I\sum_{k}a_{i}\left(\nabla B_{k0}\right)_{k} \ \ \ \ \ (14)$

For a fixed dipole, the moment is a constant and is

$\displaystyle \mathbf{m}=I\mathbf{a} \ \ \ \ \ (15)$

so the first term is

 $\displaystyle I\sum_{k}a_{k}\left(\nabla B_{k0}\right)_{i}$ $\displaystyle =$ $\displaystyle \sum_{k}m_{k}\left(\nabla B_{k0}\right)_{i}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\nabla\left(m_{k}B_{k0}\right)_{i}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla\left(\mathbf{m}\cdot\mathbf{B}_{0}\right)_{i} \ \ \ \ \ (18)$

In the second term, the sum comes out to

 $\displaystyle I\sum_{k}a_{i}\left(\nabla B_{k0}\right)_{k}$ $\displaystyle =$ $\displaystyle Ia_{i}\sum_{k}\left(\nabla B_{k0}\right)_{k}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ia_{i}\sum_{k}\frac{\partial B_{k0}}{\partial x_{k}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ia_{i}\nabla\cdot\mathbf{B}_{0}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (22)$

since ${\nabla\cdot\mathbf{B}=0}$ in general.

Thus we get the previous result

$\displaystyle \mathbf{F}=\nabla\left(\mathbf{m}\cdot\mathbf{B}\right) \ \ \ \ \ (23)$

Magnetic dipole embedded in sphere

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.27.

We’ll now consider the problem of finding the magnetic field due to an ideal dipole embedded at the centre of a sphere of linear magnetic material with permeability ${\mu}$.

This is a surprisingly tricky problem, and I must confess that I had to seek a bit of help via everyone’s friend: Google. The starting point is to try to figure out the bound currents and calculate the field from them. For a linear material, the magnetization is proportional to the auxiliary field (${\mathbf{M}=\chi_{m}\mathbf{H}}$), so the bound volume current is

$\displaystyle \mathbf{J}_{b}=\nabla\times\mathbf{M}=\nabla\times\left(\chi_{m}\mathbf{H}\right)=\chi_{m}\nabla\times\mathbf{H}=\chi_{m}\mathbf{J}_{f} \ \ \ \ \ (1)$

where ${\mathbf{J}_{f}}$ is the free volume current.

At first glance, it might seem that there is no free current in this problem, so that ${\mathbf{J}_{b}=0}$ as well. However, the dipole embedded in the sphere must be caused by a current. (Recall that for a planar loop carrying current ${I}$, the dipole moment is ${m=IA}$, where ${A}$ is the area enclosed by the loop, and an ideal dipole is just a limiting case as the size of the loop goes to zero, with the current becoming infinite in such a way as to preserve the magnitude of the magnetic moment.) Since the dipole moment is proportional to the current, we can see that the ‘free’ dipole ${\mathbf{m}}$ gives rise to a ‘bound’ dipole ${\mathbf{m}_{b}=\chi_{m}\mathbf{m}}$, also at the centre of the sphere. The effective dipole moment at the centre is therefore the sum of the free and bound dipoles, giving

$\displaystyle \mathbf{m}_{e}=\mathbf{m}+\mathbf{m}_{b}=\left(1+\chi_{m}\right)\mathbf{m}=\frac{\mu}{\mu_{0}}\mathbf{m} \ \ \ \ \ (2)$

Since the embedded dipole is the only free current in the problem, we know that ${\mathbf{J}_{b}=0}$ everywhere except at ${r=0}$, so we can work out the field due to the bound volume current at this point.

We’ve already worked out the field due to a dipole, so the field due to this effective dipole is

 $\displaystyle \mathbf{B}_{1}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi r^{3}}\left[3\left(\mathbf{m}_{e}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{m}_{e}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu}{4\pi r^{3}}\left[3\left(\mathbf{m}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{m}\right] \ \ \ \ \ (4)$

That’s the easy bit. The really tricky bit comes when we try to work out the surface bound current ${\mathbf{K}_{b}}$. As Griffiths gives the answer in the question, we can see that the field due to the surface current is supposed to be a constant times ${\mathbf{m}}$. So let’s say that the total field is

$\displaystyle \mathbf{B}=\mathbf{B}_{1}+\alpha\mathbf{m} \ \ \ \ \ (5)$

What we need in order to find ${\mathbf{K}_{b}=\mathbf{M}\times\hat{\mathbf{n}}=\mathbf{M}\times\hat{\mathbf{r}}}$ is the magnetization ${\mathbf{M}}$. Since the material is linear, we know that ${\mathbf{M}=\chi_{m}\mathbf{H}=\frac{\chi_{m}}{\mu}\mathbf{B}}$ so we get

$\displaystyle \mathbf{M}=\frac{\chi_{m}}{4\pi r^{3}}\left[3\left(\mathbf{m}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{m}\right]+\frac{\alpha\chi_{m}}{\mu}\mathbf{m} \ \ \ \ \ (6)$

Now at the surface of the sphere, ${r=R}$ so we get

 $\displaystyle \mathbf{K}_{b}$ $\displaystyle =$ $\displaystyle \mathbf{M}\times\hat{\mathbf{r}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\chi_{m}}{4\pi R^{3}}\mathbf{m}\times\hat{\mathbf{r}}+\frac{\alpha\chi_{m}}{\mu}\mathbf{m}\times\hat{\mathbf{r}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\frac{\alpha}{\mu}-\frac{1}{4\pi R^{3}}\right]\chi_{m}m\sin\theta\hat{\boldsymbol{\phi}} \ \ \ \ \ (9)$

where the last line follows because we’re taking ${\mathbf{m}=m\hat{\mathbf{z}}}$ and ${\theta}$ is the polar angle in spherical coordinates.

At this point, we have to notice that the form of ${\mathbf{K}_{b}}$ is that of a spinning shell with a constant surface charge density ${\sigma}$. This is because in that case, ${\mathbf{K}_{b}=\sigma\mathbf{v}}$, where ${\mathbf{v}}$ is the velocity of a point on the sphere, and for a sphere rotating at constant angular velocity ${\boldsymbol{\omega}}$ this is ${\mathbf{v}=\boldsymbol{\omega}\times\mathbf{r}=\omega R\sin\theta\hat{\boldsymbol{\phi}}}$. Griffiths shows in his example 5.11 that the field inside the sphere due to this surface current is

$\displaystyle \mathbf{B}=\frac{2}{3}\mu_{0}\sigma R\omega\hat{\mathbf{z}} \ \ \ \ \ (10)$

Comparing the two equations, we can make the substitution

$\displaystyle \sigma\omega R=\left[\frac{\alpha}{\mu}-\frac{1}{4\pi R^{3}}\right]\chi_{m}m \ \ \ \ \ (11)$

so the field due to ${\mathbf{K}_{b}}$ is

 $\displaystyle \mathbf{B}_{2}$ $\displaystyle =$ $\displaystyle \frac{2}{3}\mu_{0}\left[\frac{\alpha}{\mu}-\frac{1}{4\pi R^{3}}\right]\chi_{m}m\hat{\mathbf{z}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{3}\mu_{0}\left[\frac{\alpha}{\mu}-\frac{1}{4\pi R^{3}}\right]\chi_{m}\mathbf{m} \ \ \ \ \ (13)$

In order for this to match up with our original assumption 5, we must have

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{2}{3}\mu_{0}\left[\frac{\alpha}{\mu}-\frac{1}{4\pi R^{3}}\right]\chi_{m}\ \ \ \ \ (14)$ $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{2}{3}\frac{\mu_{0}\chi_{m}}{4\pi R^{3}}\left[\frac{2}{3}\frac{\mu_{0}\chi_{m}}{\mu}-1\right]^{-1} \ \ \ \ \ (15)$

We can now use ${\chi_{m}=\frac{\mu}{\mu_{0}}-1}$ to get

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{2}{3}\frac{\left(\mu-\mu_{0}\right)}{4\pi R^{3}}\left[\frac{2}{3}\left(1-\frac{\mu_{0}}{\mu}\right)-1\right]^{-1}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{3}\frac{\left(\mu-\mu_{0}\right)}{4\pi R^{3}}\left[-\frac{1}{3}-\frac{2\mu_{0}}{3\mu}\right]^{-1}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\mu\left(\mu_{0}-\mu\right)}{4\pi R^{3}\left(2\mu_{0}+\mu\right)} \ \ \ \ \ (18)$

Putting it all together, we get

$\displaystyle \mathbf{B}=\frac{\mu}{4\pi}\left[\frac{3}{r^{3}}\left(\mathbf{m}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{m}\right]+\frac{2\mu\left(\mu_{0}-\mu\right)}{4\pi R^{3}\left(2\mu_{0}+\mu\right)}\mathbf{m} \ \ \ \ \ (19)$

which is the answer given in Griffiths’s question.

Outside the sphere, the field is the sum of that due to the effective dipole at the centre and the surface bound charge. Since a spinning spherical shell behaves as an exact dipole when seen from outside, the dipole moment due to the surface charge is, using the substitution 11

 $\displaystyle \mathbf{m}_{s}$ $\displaystyle =$ $\displaystyle \frac{4}{3}\pi R^{3}\left[\frac{\alpha}{\mu}-\frac{1}{4\pi R^{3}}\right]\chi_{m}\mathbf{m}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\left[\frac{2\left(\mu_{0}-\mu\right)}{2\mu_{0}+\mu}-1\right]\left(\frac{\mu}{\mu_{0}}-1\right)\mathbf{m}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu\left(\mu_{0}-\mu\right)}{\mu_{0}\left(2\mu_{0}+\mu\right)}\mathbf{m} \ \ \ \ \ (22)$

The total dipole moment seen from outside the sphere is then

 $\displaystyle \mathbf{m}_{out}$ $\displaystyle =$ $\displaystyle \mathbf{m}_{s}+\mathbf{m}_{e}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{m}\frac{\mu}{\mu_{0}}\left(\frac{\mu_{0}-\mu}{2\mu_{0}+\mu}+1\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{m}\frac{3\mu}{2\mu_{0}+\mu} \ \ \ \ \ (25)$

The field seen outside the sphere is therefore

$\displaystyle \mathbf{B}_{out}=\frac{\mu_{0}}{4\pi r^{3}}\frac{3\mu}{2\mu_{0}+\mu}\left[3\left(\mathbf{m}\cdot\hat{\mathbf{r}}\right)\hat{\mathbf{r}}-\mathbf{m}\right] \ \ \ \ \ (26)$

Now that’s what I call a hard problem for this level of textbook. I’d advise Griffiths to mark it with the ! (more difficult than normal) symbol in the next edition 🙂