# Angular momentum conservation: example with a solenoid

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.13.

We’ll revisit the earlier problem of the two charged cylinders and the solenoid. To reiterate, we have a long solenoid with ${n}$ turns per unit length carrying current ${I_{0}}$ and a radius ${R}$, with its axis along the ${z}$ axis. The magnetic field inside the solenoid is

$\displaystyle \mathbf{B}_{0}=\mu_{0}nI\hat{\mathbf{z}} \ \ \ \ \ (1)$

The field is zero outside the solenoid.

We add two other cylinders (not solenoids), both coaxial with the solenoid. One cylinder has radius ${a (so it lies inside the solenoid) and carries surface charge ${+Q}$; the other cylinder has radius ${b>R}$ (outside the solenoid) and carries charge ${-Q}$. Both cylinders have length ${\ell}$. From Gauss’s law, the electric field between these two cylinders is, for ${a

$\displaystyle \mathbf{E}_{0}=\frac{Q}{2\pi\epsilon_{0}\ell}\frac{\hat{\mathbf{r}}}{r} \ \ \ \ \ (2)$

That is, the field points radially outward from the axis. The electric field is zero for ${r and ${r>b}$. (We’re neglecting end effects, so we’re assuming that ${\ell\gg b>a}$.)

In our earlier solution, we worked out the angular momentum contained in the fields and showed that it is equal to the mechanical angular momentum transferred to the two cylinders if the current in the solenoid is slowly reduced. However, there is another effect that we neglected: when the charged cylinders start to rotate, they generate a changing magnetic field inside them which in turn creates a circumferential electric field in the space between the cylinders. When the final rotation speeds of the two cylinders are reached (that is, when the current through the solenoid has been reduced to zero), the cylinders continue rotating, thus generating a static magnetic field that interacts with the electric field to produce an extra amount of angular momentum in the fields. We’ll consider this static magnetic field first.

The rotating cylinders are effectively solenoids themselves. The surface charge density on the two cylinders is

$\displaystyle \sigma_{a,b}=\begin{cases} \frac{Q}{2\pi a\ell} & \mbox{inner cylinder}\\ -\frac{Q}{2\pi b\ell} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (3)$

The surface current densities are

$\displaystyle K_{a,b}=\begin{cases} \frac{Q\left(a\omega_{a}\right)}{2\pi a\ell} & \mbox{inner cylinder}\\ -\frac{Q\left(b\omega_{b}\right)}{2\pi b\ell} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (4)$

so the magnetic field due to each cylinder is

$\displaystyle \mathbf{B}_{a,b}=\begin{cases} \frac{\mu_{0}Q\omega_{a}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{inner cylinder}\\ \frac{\mu_{0}Q\omega_{b}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (5)$

To get the directions of ${\mathbf{B}}$ we see from Griffiths’s Example 8.4 that the angular momentum of the inner cylinder is in the ${+z}$ direction and of the outer cylinder is in the ${-z}$ direction. Since the outer cylinder has negative charge, however, its magnetic field points in the same direction as that of the inner cylinder. Another way of putting this is to note that the angular velocities are

 $\displaystyle \boldsymbol{\omega}_{a}$ $\displaystyle =$ $\displaystyle \omega_{a}\hat{\mathbf{z}}\ \ \ \ \ (6)$ $\displaystyle \boldsymbol{\omega}_{b}$ $\displaystyle =$ $\displaystyle -\omega_{b}\hat{\mathbf{z}} \ \ \ \ \ (7)$

so the minus sign for ${\boldsymbol{\omega}_{b}}$ cancels the minus sign for the charge on the outer cylinder.

The linear momentum density is non-zero only in the region ${a\le r\le b}$ and is

 $\displaystyle \boldsymbol{\mathfrak{p}}_{em}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi^{2}\ell^{2}r}\hat{\boldsymbol{\phi}} \ \ \ \ \ (9)$

The angular momentum density is

 $\displaystyle \boldsymbol{\mathfrak{L}}_{em}$ $\displaystyle =$ $\displaystyle \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi^{2}\ell^{2}}\hat{\mathbf{z}} \ \ \ \ \ (11)$

which is constant, so the total angular momentum is

 $\displaystyle \mathbf{L}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi^{2}\ell^{2}}\left[\pi\left(b^{2}-a^{2}\right)\ell\right]\hat{\mathbf{z}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi\ell}\left(b^{2}-a^{2}\right)\hat{\mathbf{z}} \ \ \ \ \ (13)$

Now we can look at what’s happening as the cylinders are spinning up to their final speeds. As they speed up, the magnetic field due to each cylinder is changing according to

$\displaystyle \frac{\partial\mathbf{B}_{a,b}}{\partial t}=\begin{cases} \frac{\mu_{0}Q\dot{\omega}_{a}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{inner cylinder}\\ \frac{\mu_{0}Q\dot{\omega}_{b}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (14)$

where the dot indicates a time derivative. According to Faraday’s law, this changing magnetic field induces a circumferential electric field:

$\displaystyle \oint\mathbf{E}\cdot d\boldsymbol{\ell}=-\int\frac{\partial\mathbf{B}_{a,b}}{\partial t}\cdot d\mathbf{a} \ \ \ \ \ (15)$

This electric field will exert a torque on each cylinder, whose integral over time will give the angular momentum transferred to the cylinders. First we need to calculate the field at each cylinder. We choose a circular path of integration at the surface of each cylinder. Remember that the field of the inner cylinder is non-zero only for ${r, which the the field for the outer cylinder covers the entire region ${r.

 $\displaystyle \mathbf{E}_{a}$ $\displaystyle =$ $\displaystyle -\frac{1}{2\pi a}\pi a^{2}\frac{\mu_{0}Q\left(\dot{\omega}_{a}+\dot{\omega}_{b}\right)}{2\pi\ell}\hat{\boldsymbol{\phi}}\ \ \ \ \ (16)$ $\displaystyle \mathbf{E}_{b}$ $\displaystyle =$ $\displaystyle -\frac{1}{2\pi b}\frac{\mu_{0}Q\left(\pi a^{2}\dot{\omega}_{a}+\pi b^{2}\dot{\omega}_{b}\right)}{2\pi\ell}\hat{\boldsymbol{\phi}} \ \ \ \ \ (17)$

To confirm the direction of ${\mathbf{E}}$, recall Lenz’s law, which states that the induced field opposes the change that produced it. Since the magnetic field is increasing in the ${+z}$ direction, the induced electric field must oppose this increase so it must be in the ${-\phi}$ direction.

The torque on the cylinders is then

 $\displaystyle \mathbf{N}_{a}$ $\displaystyle =$ $\displaystyle \mathbf{r}\times\mathbf{F}_{a}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{r}\times Q\mathbf{E}_{a}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{2}\frac{\mu_{0}Q^{2}\left(\dot{\omega}_{a}+\dot{\omega}_{b}\right)}{4\pi\ell}\left(-\hat{\mathbf{r}}\times\hat{\boldsymbol{\phi}}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -a^{2}\frac{\mu_{0}Q^{2}\left(\dot{\omega}_{a}+\dot{\omega}_{b}\right)}{4\pi\ell}\hat{\mathbf{z}}\ \ \ \ \ (21)$ $\displaystyle \mathbf{N}_{b}$ $\displaystyle =$ $\displaystyle -\mathbf{r}\times Q\mathbf{E}_{b}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}Q^{2}\left(a^{2}\dot{\omega}_{a}+b^{2}\dot{\omega}_{b}\right)}{4\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (23)$

 $\displaystyle \mathbf{N}$ $\displaystyle =$ $\displaystyle \mathbf{N}_{a}+\mathbf{N}_{b}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}Q^{2}\left(b^{2}-a^{2}\right)\dot{\omega}_{b}}{4\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (25)$

Integrating over the time it takes to reach the final speed ${\omega_{b}}$ we get

$\displaystyle \mathbf{L}=\frac{\mu_{0}Q^{2}\left(b^{2}-a^{2}\right)\omega_{b}}{4\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (26)$

which is equal and opposite to 13. Thus the total angular momentum introduced into the system by magnetic and electric fields induced by the rotating cylinders is zero, showing that angular momentum is conserved.

# Momentum of a point charge outside a solenoid

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.14.

Another example of calculating momentum and angular momentum in electromagnetic fields. We have an infinite solenoid along the ${z}$ axis, of radius ${R}$ with ${n}$ turns per unit length and carrying current ${I}$. At position ${a\hat{\mathbf{x}}}$ there is a point charge ${q}$. We want to find the momentum and angular momentum of the resulting fields.

The first step in this problem is deciding which coordinate system to use. The solenoid makes us think of cylindrical coordinates, while the point charge suggests spherical. However, these two systems don’t mesh well, so we can try the fallback of using rectangular coordinates.

The field of the solenoid is zero outside and uniform inside, where it is

$\displaystyle \mathbf{B}=\mu_{0}nI\hat{\mathbf{z}}\mbox{ for }x^{2}+y^{2}

If the point charge were located at the origin, its field would be

$\displaystyle \mathbf{E}=\frac{q}{4\pi\epsilon_{0}}\frac{x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}}{\left(x^{2}+y^{2}+z^{2}\right)^{3/2}} \ \ \ \ \ (2)$

If we shift the charge to ${a\hat{\mathbf{x}}}$ then we get

$\displaystyle \mathbf{E}=\frac{q}{4\pi\epsilon_{0}}\frac{\left(x-a\right)\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}}{\left(\left(x-a\right)^{2}+y^{2}+z^{2}\right)^{3/2}} \ \ \ \ \ (3)$

The momentum density is

 $\displaystyle \boldsymbol{\mathfrak{p}}_{em}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{4\pi\left(\left(x-a\right)^{2}+y^{2}+z^{2}\right)^{3/2}}\left[-\left(x-a\right)\hat{\mathbf{y}}+y\hat{\mathbf{x}}\right] \ \ \ \ \ (5)$

To get the total momentum we need to integrate this over the interior of the solenoid. It turns out to be easiest to do this by integrating first over ${z}$ and then converting to polar coordinates for the remaining two integrations. Integrating over ${z}$ we get (using Maple or tables)

 $\displaystyle \frac{\mu_{0}nIq}{4\pi}\left[-\left(x-a\right)\hat{\mathbf{y}}+y\hat{\mathbf{x}}\right]\int_{-\infty}^{\infty}\frac{dz}{\left(\left(x-a\right)^{2}+y^{2}+z^{2}\right)^{3/2}}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{4\pi}\left[-\left(x-a\right)\hat{\mathbf{y}}+y\hat{\mathbf{x}}\right]\frac{2}{\left(x-a\right)^{2}+y^{2}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{2\pi}\frac{\left(a-x\right)\hat{\mathbf{y}}+y\hat{\mathbf{x}}}{\left(x-a\right)^{2}+y^{2}} \ \ \ \ \ (7)$

If we integrate the ${\hat{\mathbf{x}}}$ component over ${y}$ we have

$\displaystyle \int_{-\sqrt{R^{2}-x^{2}}}^{\sqrt{R^{2}-x^{2}}}\frac{y\hat{\mathbf{x}}}{\left(x-a\right)^{2}+y^{2}}=0 \ \ \ \ \ (8)$

because the integrand is an odd function of ${y}$ and the integral is over a symmetric interval. This leaves us with the ${\hat{\mathbf{y}}}$ component, and it is here that we turn to polar coordinates. Using ${x=r\cos\theta}$ and ${y=r\sin\theta}$ we have

 $\displaystyle \int_{0}^{R}\int_{0}^{2\pi}\frac{\left(a-x\right)\hat{\mathbf{y}}}{\left(x-a\right)^{2}+y^{2}}r\; d\theta\; dr$ $\displaystyle =$ $\displaystyle \int_{0}^{R}\int_{0}^{2\pi}\frac{\left(a-r\cos\theta\right)r\hat{\mathbf{y}}}{r^{2}-2ar\cos\theta+a^{2}}d\theta\; dr\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{y}}\int_{0}^{R}\frac{2\pi r}{a}dr\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\pi R^{2}}{a}\hat{\mathbf{y}} \ \ \ \ \ (11)$

where we used Maple to do the ${\theta}$ integral. The total momentum is thus

$\displaystyle \mathbf{P}_{em}=\frac{\mu_{0}nIq}{2\pi}\frac{\pi R^{2}}{a}\hat{\mathbf{y}}=\frac{\mu_{0}nIqR^{2}}{2a}\hat{\mathbf{y}} \ \ \ \ \ (12)$

The angular momentum density is

 $\displaystyle \boldsymbol{\mathfrak{L}}_{em}$ $\displaystyle =$ $\displaystyle \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{4\pi\left(\left(x-a\right)^{2}+y^{2}+z^{2}\right)^{3/2}}\left[x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}\right]\times\left[-\left(x-a\right)\hat{\mathbf{y}}+y\hat{\mathbf{x}}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{4\pi\left(\left(x-a\right)^{2}+y^{2}+z^{2}\right)^{3/2}}\left[z\left(x-a\right)\hat{\mathbf{x}}+yz\hat{\mathbf{y}}+\left(x\left(a-x\right)-y^{2}\right)\hat{\mathbf{z}}\right] \ \ \ \ \ (15)$

Integrating the ${\hat{\mathbf{x}}}$ and ${\hat{\mathbf{y}}}$ components over ${z}$ gives zero because the integrand is an odd function of ${z}$ integrated over a symmetric interval. Thus we are left with the ${\hat{\mathbf{z}}}$ component which we can again integrate first over ${z}$ and then over the other two coordinates using polar coordinates. We have

 $\displaystyle \mathbf{L}_{em}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{4\pi}\hat{\mathbf{z}}\int_{0}^{R}\int_{0}^{2\pi}\int_{-\infty}^{\infty}dz\frac{\left(x\left(a-x\right)-y^{2}\right)r\; d\theta\; dr}{\left(\left(x-a\right)^{2}+y^{2}+z^{2}\right)^{3/2}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{4\pi}\hat{\mathbf{z}}\int_{0}^{R}\int_{0}^{2\pi}\frac{2\left(x\left(a-x\right)-y^{2}\right)}{\left(x-a\right)^{2}+y^{2}}r\; d\theta\; dr\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{2\pi}\hat{\mathbf{z}}\int_{0}^{R}\int_{0}^{2\pi}\frac{\left(r\cos\theta\left(a-r\cos\theta\right)-r^{2}\sin^{2}\theta\right)r}{r^{2}-2ar\cos\theta+a^{2}}d\theta\; dr\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{2\pi}\hat{\mathbf{z}}\int_{0}^{R}\int_{0}^{2\pi}\frac{\left(ar\cos\theta-r^{2}\right)r}{r^{2}-2ar\cos\theta+a^{2}}d\theta\; dr\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nIq}{2\pi}\hat{\mathbf{z}}\int_{0}^{R}\left[\frac{2\pi r^{2}}{a^{2}-r^{2}}-\frac{2\pi r^{2}}{a^{2}-r^{2}}\right]r\; dr\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (21)$

where again we used Maple to do the integral.

# Energy transfer in a solenoid

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.9.

Here’s a simple example of conservation of energy in an electromagnetic system. We have an infinite solenoid of radius ${a}$ carrying ${n}$ turns per unit length and current ${I_{s}}$. The magnetic field is zero outside the solenoid and inside we have

$\displaystyle \mathbf{B}_{s}=\mu_{0}nI_{s}\hat{\mathbf{z}} \ \ \ \ \ (1)$

Now suppose we put a circular wire loop of radius ${b\gg a}$ and resistance ${R}$ around the solenoid. If we now decrease the current in the solenoid, the changing magnetic flux will induce a circumferential electric field around the solenoid, which will in turn create a current in the wire. To find the current ${I_{r}}$ in the resistor we can use the fact that the electric field creates an electromotive force (emf) around the wire:

 $\displaystyle \mathcal{E}$ $\displaystyle =$ $\displaystyle -\frac{d\Phi}{dt}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\pi a^{2}\mu_{0}n\frac{dI_{s}}{dt} \ \ \ \ \ (3)$

The current is

$\displaystyle I_{r}=\frac{\mathcal{E}}{R}=-\frac{\pi a^{2}\mu_{0}n}{R}\frac{dI_{s}}{dt} \ \ \ \ \ (4)$

To find the direction of the current, remember that it in turn generates a magnetic field that opposes the reduction in the solenoid’s field, so the current must flow in the ${+\phi}$ direction (counterclockwise as viewed from above).

The rate at which energy is dissipated by the resistor is the power, which is ${I_{r}^{2}R}$. This energy must come from the solenoid via the Poynting vector. We can calculate the Poynting vector just outside the solenoid as follows. First, we need ${\mathbf{E}}$ and ${\mathbf{B}}$ outside the solenoid. The electric field is produced by the changing magnetic field in the solenoid, and by Faraday’s law we have

$\displaystyle \oint\mathbf{E}\cdot d\boldsymbol{\ell}=-\int\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{a} \ \ \ \ \ (5)$

Taking a circular path of radius ${a}$ we get

 $\displaystyle 2\pi aE$ $\displaystyle =$ $\displaystyle -\mu_{0}n\frac{dI_{s}}{dt}\pi a^{2}\ \ \ \ \ (6)$ $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}na}{2}\frac{dI_{s}}{dt}\hat{\boldsymbol{\phi}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{I_{r}R}{2\pi a}\hat{\boldsymbol{\phi}} \ \ \ \ \ (8)$

Remember that ${\frac{dI_{s}}{dt}<0}$ so ${\mathbf{E}}$ points in the ${+\phi}$ direction.

There is no magnetic field due to the solenoid outside the solenoid itself, but the current in the resistor generates a magnetic field due to the Biot-Savart law. Griffiths works out the magnetic field on the ${z}$ axis due to a circular loop in his Example 5.6 and since we’re taking the radius ${b}$ of the loop to be much greater than the radius ${a}$ of the solenoid, we can use this formula as a good approximation. We have

$\displaystyle \mathbf{B}_{r}=\frac{\mu_{0}I_{r}}{2}\frac{b^{2}}{\left(b^{2}+z^{2}\right)^{3/2}}\hat{\mathbf{z}} \ \ \ \ \ (9)$

The Poynting vector is then

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{I_{r}^{2}R}{4\pi a}\frac{b^{2}}{\left(b^{2}+z^{2}\right)^{3/2}}\hat{\mathbf{r}} \ \ \ \ \ (11)$

We can integrate the magnitude of the vector over the surface of the solenoid to find the rate at which energy is radiating away from the solenoid. We get

 $\displaystyle P$ $\displaystyle =$ $\displaystyle \int\mathbf{S}\cdot d\mathbf{a}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(2\pi a\right)\frac{I_{r}^{2}R}{4\pi a}\int_{-\infty}^{\infty}\frac{b^{2}}{\left(b^{2}+z^{2}\right)^{3/2}}dz\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I_{r}^{2}R \ \ \ \ \ (14)$

Thus the power in the resistor is indeed coming from the solenoid.

# Angular momentum in electromagnetic fields

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 8, Post 7.

The momentum density of an electromagnetic field is given by

$\displaystyle \boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mu_{0}\mathbf{S}=\epsilon_{0}\mathbf{E}\times\mathbf{B} \ \ \ \ \ (1)$

If we have linear momentum, then we automatically have angular momentum with respect to some origin by using the classical definition of angular momentum ${\mathbf{L}=\mathbf{r}\times\mathbf{p}}$. We can define the angular momentum density of an electromagnetic field by

$\displaystyle \boldsymbol{\mathfrak{L}}_{em}\equiv\mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mathbf{r}\times\left(\mathbf{E}\times\mathbf{B}\right) \ \ \ \ \ (2)$

Just as with linear momentum, even static fields can have angular momentum. As an example, suppose we have a long solenoid with ${n}$ turns per unit length carrying current ${I_{0}}$ and a radius ${R}$, with its axis along the ${z}$ axis. The magnetic field inside the solenoid is

$\displaystyle \mathbf{B}_{0}=\mu_{0}nI\hat{\mathbf{z}} \ \ \ \ \ (3)$

The field is zero outside the solenoid.

Now suppose
we add two other cylinders (not solenoids), both coaxial with
the solenoid. One cylinder has radius ${a (so it lies inside the solenoid) and carries surface charge ${+Q}$; the other cylinder has radius ${b>R}$ (outside the solenoid) and carries charge ${-Q}$. Both cylinders have length ${\ell}$. From Gauss’s law, the electric field between these two cylinders is, for ${a

$\displaystyle \mathbf{E}_{0}=\frac{Q}{2\pi\epsilon_{0}\ell}\frac{\hat{\mathbf{r}}}{r} \ \ \ \ \ (4)$

That is, the field points radially outward from the axis. The electric field is zero for ${r and ${r>b}$. (We’re neglecting end effects, so we’re assuming that ${\ell\gg b>a}$.)

The linear momentum density is non-zero in the region ${a (where both fields are non-zero) and we have

$\displaystyle \boldsymbol{\mathfrak{p}}_{em}=-\frac{\mu_{0}nIQ}{2\pi\ell}\frac{\hat{\boldsymbol{\phi}}}{r} \ \ \ \ \ (5)$

so the angular momentum density is

 $\displaystyle \boldsymbol{\mathfrak{L}}_{em}$ $\displaystyle =$ $\displaystyle \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}nIQ}{2\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (7)$

Conveniently, this is constant so the total angular momentum is just the density times the volume of the cylindrical tube in the region ${a

 $\displaystyle \mathbf{L}_{em}$ $\displaystyle =$ $\displaystyle -\ell\pi\left(R^{2}-a^{2}\right)\frac{\mu_{0}nIQ}{2\pi\ell}\hat{\mathbf{z}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(R^{2}-a^{2}\right)\frac{\mu_{0}nIQ}{2}\hat{\mathbf{z}} \ \ \ \ \ (9)$

Now suppose we (quasistatically) discharge the two cylinders by connecting a resistor ${\mathcal{R}}$ between them. We’d like to show that the angular momentum gets transferred from the fields to the physical devices in the problem. The two cylinders are effectively a capacitor with some capacitance ${C}$, so we know that the current in the resistor will decay exponentially

$\displaystyle I\left(t\right)=\frac{V_{0}}{\mathcal{R}}e^{-t/\mathcal{R}C} \ \ \ \ \ (10)$

where ${V_{0}}$ is the potential difference between the cylinders at ${t=0}$. The force ${d\mathbf{F}}$ on a segment of the resistor of length ${dr}$ is

 $\displaystyle d\mathbf{F}$ $\displaystyle =$ $\displaystyle I\left(t\right)dr\hat{\mathbf{r}}\times\mathbf{B}_{0}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I\left(t\right)drB_{0}\hat{\boldsymbol{\phi}} \ \ \ \ \ (12)$

so the torque on this segment is

 $\displaystyle d\mathbf{N}$ $\displaystyle =$ $\displaystyle \mathbf{r}\times d\mathbf{F}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I\left(t\right)B_{0}rdr\hat{\mathbf{z}} \ \ \ \ \ (14)$

The total torque on the resistor at time ${t}$ is

 $\displaystyle \mathbf{N}\left(t\right)$ $\displaystyle =$ $\displaystyle -I\left(t\right)B_{0}\hat{\mathbf{z}}\int_{a}^{R}r\; dr\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}I\left(t\right)B_{0}\left(R^{2}-a^{2}\right)\hat{\mathbf{z}} \ \ \ \ \ (16)$

The angular impulse is the integral of torque over time, so we get

 $\displaystyle \mathbf{I}$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}\mathbf{N}\left(t\right)dt\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}B_{0}\left(R^{2}-a^{2}\right)\hat{\mathbf{z}}\int_{0}^{\infty}I\left(t\right)dt\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}B_{0}\left(R^{2}-a^{2}\right)\hat{\mathbf{z}}\int_{0}^{\infty}\frac{V_{0}}{\mathcal{R}}e^{-t/\mathcal{R}C}dt\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}B_{0}\left(R^{2}-a^{2}\right)CV_{0}\hat{\mathbf{z}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\mu_{0}nI\left(R^{2}-a^{2}\right)Q\hat{\mathbf{z}} \ \ \ \ \ (21)$

where we used the relation between capacitance, charge and voltage ${Q=CV}$. We see that this agrees with 9, so all the angular momentum in the fields is transferred to the resistor as the electric field is reduced to zero.

# Energy in a magnetic field

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 7.26.

We’ve seen that the energy stored in an electric field is

$\displaystyle W_{E}=\frac{\epsilon_{0}}{2}\int E^{2}d^{3}\mathbf{r} \ \ \ \ \ (1)$

where the integral is over all space. Here we’ll look at the derivation of a similar formula for the magnetic field.

The magnetic flux through an inductor carrying current ${I}$ is

$\displaystyle \Phi=LI \ \ \ \ \ (2)$

where ${L}$ is the inductance of the circuit. The emf induced in a circuit by changing the current in that circuit is then

$\displaystyle \mathcal{E}=-\frac{d\Phi}{dt}=-L\dot{I} \ \ \ \ \ (3)$

The power generated by a current is the voltage multiplied by the current, and this power is the rate at which work is done, so

$\displaystyle \frac{dW_{B}}{dt}=-\mathcal{E}I=LI\dot{I} \ \ \ \ \ (4)$

The sign in this equation reflects the fact that increasing the current through the inductor induces an emf ${\mathcal{E}}$ that opposes the increase, so we need to do work against this back emf to generate the current. This work should be positive for an increasing current (that is, for ${\dot{I}>0}$), so we’ve defined the signs to make this true.

If we increase the current from zero to some final value ${I}$ then by integrating this equation we get the total work done:

$\displaystyle W_{B}=\frac{1}{2}LI^{2} \ \ \ \ \ (5)$

This formula shows that the total energy delivered to the inductor depends only on the final current, and not the route by which we get there. Thus the current could increase for a while, then decrease for a bit, and then increase again, but as long as it ends up at the final value of ${I}$ the same work is required.

To convert this expression to one containing the magnetic field requires a bit of juggling with vector calculus, so here we go. First, we can write ${LI}$ in terms of ${\mathbf{B}}$:

$\displaystyle LI=\Phi=\int\mathbf{B}\cdot d\mathbf{a} \ \ \ \ \ (6)$

In terms of the vector potential, this becomes, using Stokes’s theorem:

 $\displaystyle LI$ $\displaystyle =$ $\displaystyle \int\left(\nabla\times\mathbf{A}\right)\cdot d\mathbf{a}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \oint\mathbf{A}\cdot d\boldsymbol{\ell} \ \ \ \ \ (8)$

where we’ve converted to a line integral around the circuit bordering the area. Therefore, the work is

$\displaystyle W_{B}=\frac{1}{2}I\oint\mathbf{A}\cdot d\boldsymbol{\ell}=\frac{1}{2}\oint\left(\mathbf{A}\cdot\mathbf{I}\right)d\ell \ \ \ \ \ (9)$

where we’ve made the current a vector in place of the directed line element. If the current occupies a volume rather than a linear circuit, we can write the generalization of this as

$\displaystyle W_{B}=\frac{1}{2}\int\left(\mathbf{A}\cdot\mathbf{J}\right)d^{3}\mathbf{r} \ \ \ \ \ (10)$

If these are steady currents (that is, we’ve increased the current up to a certain value and then held it there), we can apply Ampère’s law in the form ${\nabla\times\mathbf{B}=\mu_{0}\mathbf{J}}$ to get

$\displaystyle W_{B}=\frac{1}{2\mu_{0}}\int\mathbf{A}\cdot\left(\nabla\times\mathbf{B}\right)d^{3}\mathbf{r} \ \ \ \ \ (11)$

Now we can use a vector calculus identity:

$\displaystyle \nabla\cdot\left(\mathbf{A}\times\mathbf{B}\right)=\mathbf{B}\cdot\left(\nabla\times\mathbf{A}\right)-\mathbf{A}\cdot\left(\nabla\times\mathbf{B}\right) \ \ \ \ \ (12)$

to write

 $\displaystyle W_{B}$ $\displaystyle =$ $\displaystyle \frac{1}{2\mu_{0}}\int\mathbf{B}\cdot\left(\nabla\times\mathbf{A}\right)d^{3}\mathbf{r}-\frac{1}{2\mu_{0}}\int\nabla\cdot\left(\mathbf{A}\times\mathbf{B}\right)d^{3}\mathbf{r}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\mu_{0}}\int B^{2}d^{3}\mathbf{r}-\frac{1}{2\mu_{0}}\int\nabla\cdot\left(\mathbf{A}\times\mathbf{B}\right)d^{3}\mathbf{r} \ \ \ \ \ (14)$

We can use the divergence theorem to convert the second integral to a surface integral, and then perform the usual trick of letting the surface go to infinity. We get

$\displaystyle W_{B}=\frac{1}{2\mu_{0}}\int B^{2}d^{3}\mathbf{r}-\frac{1}{2\mu_{0}}\int\left(\mathbf{A}\times\mathbf{B}\right)\cdot d\mathbf{a} \ \ \ \ \ (15)$

If the currents are all localized, then both ${\mathbf{A}}$ and ${\mathbf{B}}$ tend to zero at infinity, so we can ignore this final integral and get

$\displaystyle W_{B}=\frac{1}{2\mu_{0}}\int B^{2}d^{3}\mathbf{r} \ \ \ \ \ (16)$

This is the energy stored in a (localized) magnetic field produced by steady currents.

As an example, we’ll consider the standard case of the infinite solenoid (I know, I know, we derived this formula for finite current distributions, so you can think of this problem as a ‘very long’ solenoid rather than an infinite one), with ${n}$ turns per unit length carrying current ${I}$. We’ll work out the energy per unit length of a section far from the ends.

If the solenoid has radius ${R}$, its inductance per unit length is

$\displaystyle L=\pi R^{2}\mu_{0}n^{2} \ \ \ \ \ (17)$

so from 5 we have

$\displaystyle W_{B}=\frac{1}{2}\pi R^{2}\mu_{0}n^{2}I^{2} \ \ \ \ \ (18)$

We can also work this out from 9 using the vector potential of the solenoid given by Griffiths in his example 5.12:

$\displaystyle \mathbf{A}=\frac{\mu_{0}nIr}{2}\hat{\boldsymbol{\phi}} \ \ \ \ \ (19)$

for ${r.

In a solenoid, ${\mathbf{I}}$ is in the ${\phi}$ direction so the total current in unit length is ${nI\hat{\boldsymbol{\phi}}}$, the line integral goes around the solenoid at ${r=R}$ and we get

$\displaystyle \frac{1}{2}\oint\left(\mathbf{A}\cdot\mathbf{I}\right)d\ell=\frac{1}{2}\frac{\mu_{0}nIR}{2}2\pi RnI=\frac{1}{2}\pi R^{2}\mu_{0}n^{2}I^{2} \ \ \ \ \ (20)$

Next, we can use 16. The field inside the solenoid is a constant

$\displaystyle B=n\mu_{0}I \ \ \ \ \ (21)$

and the volume per unit length is ${\pi R^{2}}$ so

$\displaystyle W_{B}=\frac{1}{2\mu_{0}}\pi R^{2}\left(n\mu_{0}I\right)^{2}=\frac{1}{2}\pi R^{2}\mu_{0}n^{2}I^{2} \ \ \ \ \ (22)$

Finally, we can use 15. Instead of taking the integration volume to be all space, we can use any volume that completely encloses the current, so we can use a cylindrical tube of inner radius ${r=a to ${r=b>R}$. Outside the solenoid, ${\mathbf{B}=0}$ so we need look only at the region ${a\le r\le R}$. The first integral is

$\displaystyle \frac{1}{2\mu_{0}}\int_{a\le r\le R}B^{2}d^{3}\mathbf{r}=\frac{1}{2}\pi\left(R^{2}-a^{2}\right)\mu_{0}n^{2}I^{2} \ \ \ \ \ (23)$

To do the surface integral, we first work out the direction of ${\mathbf{A}\times\mathbf{B}}$. ${\mathbf{A}}$ is in the ${\phi}$ direction and ${\mathbf{B}}$ is in the ${z}$ direction, so ${\mathbf{A}\times\mathbf{B}}$ is in the radial direction, pointing outwards. On the inner surface of the tube, ${d\mathbf{a}}$ points radially inwards, so ${\left(\mathbf{A}\times\mathbf{B}\right)\cdot d\mathbf{a}<0}$, and the integral is

$\displaystyle -\frac{1}{2\mu_{0}}\int_{r=a}\left(\mathbf{A}\times\mathbf{B}\right)\cdot d\mathbf{a}=\frac{1}{2\mu_{0}}2\pi a\frac{\mu_{0}nIa}{2}n\mu_{0}I=\frac{1}{2}\pi a^{2}\mu_{0}n^{2}I^{2} \ \ \ \ \ (24)$

Adding the two contributions, we get

$\displaystyle W_{B}=\frac{1}{2}\pi\left(R^{2}-a^{2}\right)\mu_{0}n^{2}I^{2}+\frac{1}{2}\pi a^{2}\mu_{0}n^{2}I^{2}=\frac{1}{2}\pi R^{2}\mu_{0}n^{2}I^{2} \ \ \ \ \ (25)$

Thus all four methods give the same answer.

# Auxiliary magnetic field H

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.12.

Jackson, J. D. (1999) Classical Electrodynamics, 3rd Edition; Wiley – Sections 5.8, 5.9.

An object containing a magnetization distribution can be modelled by replacing the magnetization by bound volume and surface currents ${\mathbf{J}_{b}}$ and ${\mathbf{K}_{b}}$. If we add in some extra free current ${\mathbf{J}_{f}}$ not due to the magnetization (for example, by plugging the object into the electric mains), then the total current at a point inside the object is

$\displaystyle \mathbf{J}=\mathbf{J}_{b}+\mathbf{J}_{f} \ \ \ \ \ (1)$

Since ${\mathbf{J}_{b}=\nabla\times\mathbf{M}}$ by definition, we can write Ampère’s law as

 $\displaystyle \frac{1}{\mu_{0}}\nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mathbf{J}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{J}_{f}+\nabla\times\mathbf{M}\ \ \ \ \ (3)$ $\displaystyle \nabla\times\left(\frac{1}{\mu_{0}}\mathbf{B}-\mathbf{M}\right)$ $\displaystyle =$ $\displaystyle \mathbf{J}_{f} \ \ \ \ \ (4)$

The quantity in parentheses is given the symbol ${\mathbf{H}}$:

$\displaystyle \mathbf{H}\equiv\frac{1}{\mu_{0}}\mathbf{B}-\mathbf{M} \ \ \ \ \ (5)$

${\mathbf{H}}$ is sometimes called the auxiliary magnetic field or sometimes just the magnetic field, with ${\mathbf{B}}$ being called the magnetic flux density. This allows a variant form of Ampère’s law:

 $\displaystyle \nabla\times\mathbf{H}$ $\displaystyle =$ $\displaystyle \mathbf{J}_{f}\ \ \ \ \ (6)$ $\displaystyle \oint\mathbf{H}\cdot d\boldsymbol{\ell}$ $\displaystyle =$ $\displaystyle I_{f} \ \ \ \ \ (7)$

where ${I_{f}}$ is the free current enclosed by the path of integration in the second line.

One thing that is a bit worrying is that the bound surface current ${\mathbf{K}_{b}}$ seems to have vanished in this derivation. Griffiths makes no mention of this, but Jackson gets round the problem by saying that the surface integral from which ${\mathbf{K}_{b}}$ was defined, namely ${\frac{\mu_{0}}{4\pi}\int\frac{\mathbf{M}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\times d\mathbf{a}^{\prime}}$ is zero by assuming that the magnetization ${\mathbf{M}}$ is well-behaved and localized, and we can take the surface at infinity where ${\mathbf{M}=0}$. He then states later that in some idealized problems, it is convenient to assume that ${\mathbf{M}}$ is discontinuous at the boundary between two objects (for example, between a magnetized object and the surrounding air), and in that case, the surface current term must be added in. However, in any ‘real’ physical situation, discontinuities never occur so the surface term doesn’t appear and the definition of ${\mathbf{H}}$ above is valid.

In any case, we can use this definition of ${\mathbf{H}}$ to calculate ${\mathbf{B}}$ more easily in some idealized situations. For example, if we have an infinitely long cylinder of radius ${R}$ with a fixed ${\mathbf{M}=kr\hat{\mathbf{z}}}$, we can find ${\mathbf{B}}$ by two methods.

First, we use the bound current approach.

 $\displaystyle \mathbf{J}_{b}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{M}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -k\hat{\boldsymbol{\phi}}\ \ \ \ \ (9)$ $\displaystyle \mathbf{K}_{b}$ $\displaystyle =$ $\displaystyle \mathbf{M}\left(R\right)\times\hat{\mathbf{n}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle kR\hat{\boldsymbol{\phi}} \ \ \ \ \ (11)$

Note that the total bound current is zero, since the total volume current is

$\displaystyle \int_{0}^{R}\mathbf{J}_{b}dr=-kR\hat{\boldsymbol{\phi}} \ \ \ \ \ (12)$

Both bound currents effectively produce solenoids, so the field outside the cylinder is zero. Inside, we have field due to the surface current, which is

$\displaystyle \mathbf{B}_{K}=\mu_{0}kR\hat{\mathbf{z}} \ \ \ \ \ (13)$

and we must add to that the field due to those parts of the cylinder with a radius greater than the the radius ${r}$ of interest. The total current outside radius ${r}$ is

$\displaystyle \int_{r}^{R}\mathbf{J}_{b}dr=-k\left(R-r\right)\hat{\boldsymbol{\phi}} \ \ \ \ \ (14)$

so the total field is

$\displaystyle \mathbf{B}=\mu_{0}kR\hat{\mathbf{z}}-\mu_{0}k\left(R-r\right)\hat{\mathbf{z}}=\mu_{0}kr\hat{\mathbf{z}}=\mu_{0}\mathbf{M} \ \ \ \ \ (15)$

Using ${\mathbf{H}}$, we can take a loop of integration of radius ${r}$ centred on the ${z}$ axis. Since there is no free current, we get

$\displaystyle \oint\mathbf{H}\cdot d\boldsymbol{\ell}=0 \ \ \ \ \ (16)$

and from the symmetry of the problem we can conclude that ${\mathbf{H}=0}$.

Alternatively, we can work from the curl equation which gives ${\nabla\times\mathbf{H}=0}$. This on its own isn’t enough to conclude that ${\mathbf{H}=0}$, but we can also calculate the divergence

$\displaystyle \nabla\cdot\mathbf{H}=\frac{1}{\mu_{0}}\nabla\cdot\mathbf{B}-\nabla\cdot\mathbf{M} \ \ \ \ \ (17)$

Since ${\nabla\cdot\mathbf{B}=0}$ always, and by direct calculation we can show that ${\nabla\cdot\mathbf{M}=0}$ in this case, we have both the curl and divergence of ${\mathbf{H}}$ as zero, so ${\mathbf{H}}$ must be zero. From that, we can conclude immediately from the definition of ${\mathbf{H}}$ that ${\mathbf{B}=\mu_{0}\mathbf{M}}$.

# Toroidal solenoid with a gap

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.10.

As another example of how we can calculate the magnetic field due to the magnetization of an object, suppose we start with an iron bar with square cross-section of side length ${a}$ and length ${L}$, and having a constant magnetization ${\mathbf{M}}$ parallel to the length of the bar. We now bend the bar into a circle so that the end points almost touch, but are separated by a gap of width ${w}$. Our task is to find the magnetic field at the centre of the gap.

The first thing to recognize is that by bending the bar into a circle, the magnetization is no longer constant in magnitude within the bar, since the outer edge of the circle will be stretched relative to the inside edge. The new magnetization will be

$\displaystyle M^{\prime}\left(r\right)=\frac{L}{2\pi r}M \ \ \ \ \ (1)$

We can verify this by finding the magnitude of the total dipole moment in the bar:

$\displaystyle \int_{L/2\pi}^{L/2\pi+a}2\pi raM^{\prime}\left(r\right)dr=a^{2}LM \ \ \ \ \ (2)$

so the total magnetic moment is the same as in the original unbent bar. Since the magnetization now points along the circle, its vector form is

$\displaystyle \mathbf{M}^{\prime}=\frac{L}{2\pi r}M\hat{\boldsymbol{\phi}} \ \ \ \ \ (3)$

From the formula for the curl in cylindrical coordinates, we then get for the bound volume current

$\displaystyle \mathbf{J}_{b}=\nabla\times\mathbf{\mathbf{M}^{\prime}}=0 \ \ \ \ \ (4)$

The surface current has the values on the four sides of the square:

$\displaystyle \mathbf{K}_{b}=\begin{cases} M\hat{\mathbf{z}} & \mbox{inside edge}\\ \frac{LM}{2\pi r}\hat{\mathbf{r}} & \mbox{top edge}\\ -\frac{ML}{\left(L+2\pi a\right)}\hat{\mathbf{z}} & \mbox{outside edge}\\ -\frac{LM}{2\pi r}\hat{\mathbf{r}} & \mbox{bottom edge} \end{cases} \ \ \ \ \ (5)$

If we assume that ${L\gg a}$ so that ${r\approx L/2\pi}$ for the whole width of the rod, we can approximate ${\mathbf{K}_{b}}$ as a current density of magnitude ${M}$ on all four sides, so we have essentially a torus-shaped solenoid with a square cross section. Griffiths works out the field inside a toroidal solenoid of arbitrary cross-section as Example 5.10, using methods similar to those we used for a linear solenoid, with the result (translating to the quantities used in this post):

 $\displaystyle \mathbf{B}_{torus}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}ML}{2\pi\left(L/2\pi\right)}\hat{\boldsymbol{\phi}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mu_{0}M\hat{\boldsymbol{\phi}} \ \ \ \ \ (7)$

This is the field inside a complete torus. To handle the gap, we can approximate it by a square current loop with current opposing that in the torus. We’ve worked out the field at the centre of such a loop before, so again translating this into the quantities used here, we have:

$\displaystyle \mathbf{B}_{loop}=-\frac{2\sqrt{2}Mw\mu_{0}}{\pi a}\hat{\boldsymbol{\phi}} \ \ \ \ \ (8)$

The net field at the centre of the gap is the sum of these two:

$\displaystyle \mathbf{B}=\mu_{0}M\hat{\boldsymbol{\phi}}\left(1-\frac{2\sqrt{2}w}{\pi a}\right) \ \ \ \ \ (9)$

# Magnetization: bound currents

References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 6.7, 6.8, 6.9.

By analogy with the polarization in electrostatics, we can define the magnetization ${\mathbf{M}}$, which is the magnetic dipole moment per unit volume. We can get a formula for the vector potential (and hence the magnetic field, although there are often easier ways of finding the field) of an object containing a given magnetization (which is a vector field). Starting with the vector potential of an ideal dipole at the origin:

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi r^{2}}\mathbf{m}\times\hat{\mathbf{r}} \ \ \ \ \ (1)$

we can write this more generally as the potential when the dipole is at position ${\mathbf{r}^{\prime}}$:

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}\mathbf{m}\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \ \ \ \ \ (2)$

Then if ${\mathbf{M}=\mathbf{M}\left(\mathbf{r}^{\prime}\right)}$ we can get the potential due to a distribution of magnetic dipoles as

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}\mathbf{M}\left(\mathbf{r}^{\prime}\right)\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (3)$

For pretty well any configuration, this integral is difficult or impossible to calculate analytically, but we can transform it into a different form, in a similar way to that used in the electrostatic case for polarization. First, we use the formula

$\displaystyle \nabla^{\prime}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=\frac{\mathbf{r}-\mathbf{r}^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}} \ \ \ \ \ (4)$

so the potential becomes

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi}\int\mathbf{M}\left(\mathbf{r}^{\prime}\right)\times\nabla^{\prime}\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (5)$

Now we can use a vector product rule:

$\displaystyle \nabla\times\left(f\mathbf{V}\right)=f\left(\nabla\times\mathbf{V}\right)-\mathbf{V}\times\nabla f \ \ \ \ \ (6)$

With ${f=1/\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}$ and ${\mathbf{V}=\mathbf{M}}$ we get

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\nabla\times\mathbf{M}d^{3}\mathbf{r}^{\prime}-\frac{\mu_{0}}{4\pi}\int\nabla\times\left(\frac{\mathbf{M}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (7)$

The first integral looks like the potential of a volume current density

 $\displaystyle \mathbf{J}_{b}$ $\displaystyle \equiv$ $\displaystyle \nabla\times\mathbf{M}\ \ \ \ \ (8)$ $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\frac{\mathbf{J}_{b}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime}-\frac{\mu_{0}}{4\pi}\int\nabla\times\left(\frac{\mathbf{M}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)d^{3}\mathbf{r}^{\prime} \ \ \ \ \ (9)$

The second integral can be transformed into a surface integral by using the divergence theorem. For a general vector field ${\mathbf{V}}$ and a constant vector field ${\mathbf{c}}$ we have, using a vector identity in the first line:

 $\displaystyle \int\nabla\cdot\left(\mathbf{V}\times\mathbf{c}\right)d^{3}\mathbf{r}^{\prime}$ $\displaystyle =$ $\displaystyle \int\mathbf{V}\cdot\left(\nabla\times\mathbf{c}\right)d^{3}\mathbf{r}^{\prime}-\int\mathbf{c}\cdot\left(\nabla\times\mathbf{V}\right)d^{3}\mathbf{r}^{\prime}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int\mathbf{c}\cdot\left(\nabla\times\mathbf{V}\right)d^{3}\mathbf{r}^{\prime}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mathbf{c}\cdot\int\left(\nabla\times\mathbf{V}\right)d^{3}\mathbf{r}^{\prime}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\left(\mathbf{V}\times\mathbf{c}\right)\cdot d\mathbf{a}^{\prime}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{c}\cdot\int\mathbf{V}\times d\mathbf{a}^{\prime} \ \ \ \ \ (14)$

Thus

$\displaystyle -\int\left(\nabla\times\mathbf{V}\right)d^{3}\mathbf{r}^{\prime}=\int\mathbf{V}\times d\mathbf{a}^{\prime} \ \ \ \ \ (15)$

so

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi}\int\frac{\mathbf{J}_{b}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime}+\frac{\mu_{0}}{4\pi}\int\frac{\mathbf{M}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\times d\mathbf{a}^{\prime} \ \ \ \ \ (16)$

If we now define a surface current

$\displaystyle \mathbf{K}_{b}\equiv\mathbf{M}\times\hat{\mathbf{n}} \ \ \ \ \ (17)$

where ${\hat{\mathbf{n}}}$ is the unit normal to the surface, we get

$\displaystyle \mathbf{A}=\frac{\mu_{0}}{4\pi}\int\frac{\mathbf{J}_{b}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}d^{3}\mathbf{r}^{\prime}+\frac{\mu_{0}}{4\pi}\int\frac{\mathbf{K}_{b}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}da^{\prime} \ \ \ \ \ (18)$

That is, we can replace the volume magnetization by a volume bound current ${\mathbf{J}_{b}}$ and a surface bound current ${\mathbf{K}_{b}}$ and use them to calculate the potential and the field.

In situations where there is some symmetry, we can use Ampère’s law to calculate the field from these bound currents, as this usually proves a lot easier than trying to do the integrals.

Here are a few examples.

Example 1 First, suppose we have an infinite circular cylinder containing a uniform magnetization ${\mathbf{M}}$ parallel to its axis. Since ${\mathbf{M}}$ is constant, ${\mathbf{J}_{b}=0}$ and

$\displaystyle \mathbf{K}_{b}=M\hat{\boldsymbol{\phi}} \ \ \ \ \ (19)$

That is, the surface current flows in a circle around the outside of the cylinder. This is essentially the same as an infinite solenoid, so we know that the field outside the cylinder is zero, and inside, we have ${B=\mu_{0}nI}$ where ${n}$ is the number of turns per unit length and ${I}$ is the current, so ${nI=K_{b}=M}$. The direction of the field is given by the right hand rule, which means it’s pointing in the same direction as ${\mathbf{M}}$, thus inside:

$\displaystyle \mathbf{B}=\mu_{0}\mathbf{M} \ \ \ \ \ (20)$

Example 2 An infinite circular cylinder of radius ${R}$ has magnetization ${\mathbf{M}=kr^{2}\hat{\boldsymbol{\phi}}}$ for a constant ${k}$. The bound currents are

 $\displaystyle \mathbf{K}_{b}$ $\displaystyle =$ $\displaystyle \mathbf{M}\times\hat{\mathbf{n}}=-kR^{2}\hat{\mathbf{z}}\ \ \ \ \ (21)$ $\displaystyle \mathbf{J}_{b}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{M}=\frac{1}{r}\frac{\partial}{\partial r}\left(r\times kr^{2}\right)\hat{\mathbf{z}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3kr\hat{\mathbf{z}} \ \ \ \ \ (23)$

Inside the cylinder, using Ampère’s law the field due to the volume current (there is no contribution from the surface current) is

 $\displaystyle 2\pi rB_{J}$ $\displaystyle =$ $\displaystyle 2\pi\mu_{0}\left(3k\right)\int_{0}^{r}\left(r'\right)^{2}dr'\ \ \ \ \ (24)$ $\displaystyle \mathbf{B}_{J}$ $\displaystyle =$ $\displaystyle \mu_{0}kr^{2}\hat{\boldsymbol{\phi}} \ \ \ \ \ (25)$

Outside, the field due to the volume current is obtained from Ampère’s law by integrating around a circle of radius ${r}$. The volume current stops at a radius ${R, so we have

$\displaystyle \mathbf{B}_{J}=\mu_{0}k\frac{R^{3}}{r}\hat{\boldsymbol{\phi}} \ \ \ \ \ (26)$

and the field due to the surface current is

 $\displaystyle 2\pi rB_{K}$ $\displaystyle =$ $\displaystyle \mu_{0}\left(2\pi R\right)kR^{2}\ \ \ \ \ (27)$ $\displaystyle \mathbf{B}_{K}$ $\displaystyle =$ $\displaystyle -\mu_{0}k\frac{R^{3}}{r}\hat{\boldsymbol{\phi}} \ \ \ \ \ (28)$

In the first line, the total surface current is obtained by multiplying ${K_{b}}$ by the circumference of the cylinder, which is ${2\pi R}$.

Thus the total field is

$\displaystyle \mathbf{B}=\mathbf{B}_{J}+\mathbf{B}_{K}=0 \ \ \ \ \ (29)$

Note that the total current (volume + surface) is zero, since they are equal in magnitude but opposite in direction. Thus by Ampère’s law, the field outside the cylinder must be zero since the enclosed current is zero.

Example 3 Finally, for a cylinder with finite length and a constant ${\mathbf{M}}$ parallel to its axis, ${\mathbf{J}_{B}=0}$ and ${\mathbf{K}_{B}=M\hat{\boldsymbol{\phi}}}$ as before, but ${\mathbf{K}_{B}=0}$ on the ends of the cylinder. The magnetic field lines come out of the north pole end of the cylinder and loop around to go back into the south pole end. For a long narrow cylinder, we’re back to the infinite solenoid, while for a short wide cylinder we have essentially a planar current loop.

# Magnetic dipole field of a finite solenoid

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.61.

For our last post on magnetostatics, we’ll consider a finite solenoid of radius ${R}$ and length ${L}$, with a surface charge density ${\sigma}$ rotating at angular speed ${\omega}$. We know that the field outside an infinite solenoid is zero, but what about a finite solenoid? For points far from the axis, we can use a dipole approximation.

We align the axis along the ${z}$ axis, and consider it to be a stack of individual current loops, each with its own dipole moment. The moment of a current loop is

$\displaystyle \mathbf{m}=\pi IR^{2}\hat{\mathbf{z}} \ \ \ \ \ (1)$

In terms of the parameters of the problem, each loop has a thickness of ${dz}$ and thus carries a current of ${I=\sigma R\omega dz}$. The contribution of the loop at coordinate ${z}$ is therefore

$\displaystyle d\mathbf{m}=\pi\omega\sigma R^{3}dz\hat{\mathbf{z}} \ \ \ \ \ (2)$

The field of the dipole from this current loop is

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

We now need to consider carefully what is meant by ${\hat{\mathbf{r}}}$. We’ll take the observation point to be a distance ${s}$ along a line perpendicular to the axis and intersecting the axis at its midpoint. We’ll define ${s}$ to be on the ${\hat{\mathbf{x}}}$ axis. Then for a given current loop at coordinate ${z}$, the vector ${\mathbf{r}}$ points from the centre of this loop to ${s}$. Therefore

 $\displaystyle r$ $\displaystyle =$ $\displaystyle \sqrt{s^{2}+z^{2}}\ \ \ \ \ (4)$ $\displaystyle \cos\theta$ $\displaystyle =$ $\displaystyle -\frac{z}{r}\ \ \ \ \ (5)$ $\displaystyle \sin\theta$ $\displaystyle =$ $\displaystyle \frac{s}{r}\ \ \ \ \ (6)$ $\displaystyle d\mathbf{m}\cdot\hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \pi\omega\sigma R^{3}\cos\theta dz\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\pi\omega\sigma R^{3}\frac{z}{r}dz \ \ \ \ \ (8)$

where ${\theta}$ is, as usual, the angle between ${\hat{\mathbf{r}}}$ and ${\hat{\mathbf{z}}}$. Therefore

 $\displaystyle \hat{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \cos\theta\hat{\mathbf{z}}+\sin\theta\hat{\mathbf{x}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{z}{r}\hat{\mathbf{z}}+\frac{s}{r}\hat{\mathbf{x}} \ \ \ \ \ (10)$

The total dipole field of the solenoid is therefore

 $\displaystyle \mathbf{B}_{dip}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\pi\omega\sigma R^{3}\int_{-\frac{L}{2}}^{\frac{L}{2}}\left[\left(\frac{3z^{2}}{\left(s^{2}+z^{2}\right)^{5/2}}-\frac{1}{\left(s^{2}+z^{2}\right)^{3/2}}\right)\hat{\mathbf{z}}+\frac{sz}{\left(s^{2}+z^{2}\right)^{5/2}}\hat{\mathbf{x}}\right]dz\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega\sigma R^{3}L}{4\left(s^{2}+\left(\frac{L}{2}\right)^{2}\right)^{3/2}}\hat{\mathbf{z}}+0\hat{\mathbf{x}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}\omega\sigma R^{3}L}{4\left(s^{2}+\left(\frac{L}{2}\right)^{2}\right)^{3/2}}\hat{\mathbf{z}} \ \ \ \ \ (13)$

Note that as ${L\rightarrow\infty}$, the field does tend to zero as ${\frac{1}{L^{2}}}$ which is the correct value for an infinite solenoid.

# Solenoid field from Biot-Savart law

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.44.

We’ve used Ampère’s law to find the field inside and outside a solenoid. The fundamental formula for finding the magnetic field due to a current is the Biot-Savart law, so it should be possible to work out the solenoid field from that as well. Since we treat the current in a solenoid as cylindrical surface current, the form of the Biot-Savart law to use is

$\displaystyle \mathbf{B}=\frac{\mu_{0}}{4\pi}\int\frac{\mathbf{K}\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}da^{\prime} \ \ \ \ \ (1)$

where ${\mathbf{K}}$ is the surface current density. For a solenoid with ${n}$ turns per unit length carrying a current ${I}$, ${K=nI}$.

Although the natural coordinates to use are cylindrical, I find it easier to set the problem up in rectangular coordinates and then convert to cylindrical later on. To define the problem, put the axis of the solenoid on the ${z}$ axis and place the observation point ${\mathbf{r}}$ on the ${x}$ axis. The source point ${\mathbf{r}^{\prime}}$ lies on the solenoid. If the radius of the solenoid is ${R}$, then

 $\displaystyle \mathbf{r}^{\prime}$ $\displaystyle =$ $\displaystyle x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}\ \ \ \ \ (2)$ $\displaystyle x^{2}+y^{2}$ $\displaystyle =$ $\displaystyle R^{2}\ \ \ \ \ (3)$ $\displaystyle \mathbf{r}$ $\displaystyle =$ $\displaystyle r\hat{\mathbf{x}}\ \ \ \ \ (4)$ $\displaystyle \mathbf{r}-\mathbf{r}^{\prime}$ $\displaystyle =$ $\displaystyle \left(r-x\right)\hat{\mathbf{x}}-y\hat{\mathbf{y}}-z\hat{\mathbf{z}} \ \ \ \ \ (5)$

Since ${\mathbf{K}}$ points around the circumference of the solenoid,

 $\displaystyle \mathbf{K}$ $\displaystyle =$ $\displaystyle nI\hat{\mathbf{\boldsymbol{\phi}}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle nI\left(-\sin\phi\hat{\mathbf{x}}+\cos\phi\hat{\mathbf{y}}\right)\ \ \ \ \ (7)$ $\displaystyle \mathbf{K}\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$ $\displaystyle =$ $\displaystyle nI\left[-z\cos\phi\hat{\mathbf{x}}-z\sin\phi\hat{\mathbf{y}}+\left(y\sin\phi-\left(r-x\right)\cos\phi\right)\hat{\mathbf{z}}\right] \ \ \ \ \ (8)$

From the symmetry of the setup, if we replace ${z}$ by ${-z}$ in the source point ${\mathbf{r}^{\prime}}$, ${\mathbf{K}}$ remains unchanged, but the ${x}$ and ${y}$ components of ${\mathbf{K}\times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}$ change sign. Thus these components cancel out and the net field must lie in the ${z}$ direction, so we can restrict our attention to that from now on.

We can now make the conversion to cylindrical coordinates using

 $\displaystyle y$ $\displaystyle =$ $\displaystyle R\sin\phi\ \ \ \ \ (9)$ $\displaystyle x$ $\displaystyle =$ $\displaystyle R\cos\phi\ \ \ \ \ (10)$ $\displaystyle da^{\prime}$ $\displaystyle =$ $\displaystyle Rd\phi dz \ \ \ \ \ (11)$

We get

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nI}{4\pi}\hat{\mathbf{z}}\int_{0}^{2\pi}\int_{-\infty}^{\infty}\frac{R\left(\sin^{2}\phi+\cos^{2}\phi\right)-r\cos\phi}{\left[\left(r-x\right)^{2}+y^{2}+z^{2}\right]^{3/2}}Rdzd\phi\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}nI}{4\pi}\hat{\mathbf{z}}\int_{0}^{2\pi}\int_{-\infty}^{\infty}\frac{R^{2}-rR\cos\phi}{\left[R^{2}+r^{2}-2rR\cos\phi+z^{2}\right]^{3/2}}dzd\phi \ \ \ \ \ (13)$

At this point it’s important to do the integrals in the right order. Attempting to do the ${\phi}$ integral first leads to a mess containing elliptic functions. If we do the ${z}$ integral first, we get

$\displaystyle \mathbf{B}=\frac{2\mu_{0}nI}{4\pi}\hat{\mathbf{z}}\int_{0}^{2\pi}\frac{R^{2}-rR\cos\phi}{R^{2}+r^{2}-2rR\cos\phi}d\phi \ \ \ \ \ (14)$

Using software, this integral comes out to

$\displaystyle \mathbf{B}=\begin{cases} \mu_{0}nI\hat{\mathbf{z}} & rR \end{cases} \ \ \ \ \ (15)$

This reproduces, after a lot of effort, the result obtained from Ampère’s law.