- #1

WeiShan Ng

- 36

- 2

## Homework Statement

A point charge q sits at the origin. A magnetic field ##\mathbf{B} (\mathbf{r})=B(x,y)\mathbf{\hat{z}}## fills all of space. The problem asks us to write down an expression for the total electromagnetic field angular momentum ##\bf{L_{EM}}##, in terms of q and the magnetic flux, ##\Phi_B## through the xy-plane

## Homework Equations

momentum density

$$\mathbf{g} = \epsilon_0 (\mathbf{E} \times \mathbf{B})$$

angular momentum density:

$$l = \vec{r} \times \vec{g} = \epsilon_0 [\vec{r}\times (\vec{E}\times \vec{B})]$$

## The Attempt at a Solution

The electric field from q is

$$\vec{E}=\frac{Q}{4\pi \epsilon_0 r^2} \mathbf{\hat{r}}$$

The magnetic field is

$$\vec{B}=B(x,y) \mathbf{\hat{z}}$$

So the angular momentum density is

$$\begin{align*} \mathbf{l} &= \epsilon_0 [\vec{r} \times (\mathbf{E}\times \mathbf{B})] \\

&=\epsilon_0 r\frac{QB}{4\pi \epsilon_0 r^2} [\mathbf{\hat{r} \times \hat{r} \times \mathbf{\hat{z}}}] \end{align*} $$

and

$$[\mathbf{\hat{r}\times \hat{r} \times \hat{z}}] = -r \sin \theta \cos \theta \cos \phi \mathbf{\hat{x}}+ r \sin \theta \cos \theta \sin \phi \mathbf{\hat{y}} + -r \sin^2 \theta \mathbf{\hat{z}}$$

The total angular momentum will be

$$\mathbf{L} = \iiint \mathbf{l} \, r^2 \sin \theta \, dr d\theta d\phi$$

Since we know that ##\int \sin \phi \, d\phi ## and ##\int \cos \phi \, d\phi## equal to zero in the interval ##[0, 2\pi]##, we can disregard x and y component, so the total angular momentum will be

$$\begin{align*} \mathbf{L} &= \iiint \frac{QB}{4\pi r} (-r \sin^2 \theta ) \, r^2 \sin \theta \, dr d\theta d\phi \, \mathbf{\hat{z}} \\ &= \iiint -\frac{QBr^2}{4\pi} \sin^3 \theta \, dr d\theta d\phi \end{align*}$$

Since we are considering xy-plane, (The surface element in a surface of polar angle θ constant) the magnetic flux can be written as

$$d\Phi = BdA = B r \sin \theta d\phi dr$$

$$\Phi = \iint B r \sin \theta d\phi dr$$

My problem is how can I substitute ##\Phi## into the equation of ##\mathbf{L}##, since I will have an extra ##r \sin^2 \theta ## in my integral ?

$$\mathbf{L} = \iiint -(B r \sin \theta) \frac{Qr\sin^2 \theta}{4\pi} \, dr d\theta d\phi \, \mathbf{\hat{z}}$$