Calculate electromagnetic momentum given a solenoid and point charge

[Bhumble]

Homework Statement

A point charge q is a distance a > R from the axis of an infinite solenoid (radius R, n turns per unit length, current I). Find the linear momentum and the angular momentum in the fields. (Put q on the x axis, with the solenoid along z; treat the solenoid as a nonconductor, so that you don’t need to worry about the induced charges on its surface).
[Answer: $$\vec{p_{em}} = \frac{\mu_0 q n I R^2}{2a} and \vec{L_{em}} = 0$$

Homework Equations

$$\vec{\rho_{em}} = \epsilon_0 \vec{E}\times\vec{B}$$

The Attempt at a Solution

So I know that I'm suppose to get the linear momentum density and then integrate over surface area to get the momentum.
And for linear momentum density I just take r X (E x B) then integrate over the surface area.
I have $$B(s>R) = 0 and B(s<R) = \mu_0 n I \hat{z}$$
The problem I'm having is setting up the electric field.
I know that it is $$\vec{E} = \frac{q}{4 \pi \epsilon_0 r^2} \hat{r}$$

r^2 = (x-a)^2 + y^2 + z^2 and since there is translational invariance along $$\hat{z}$$ I just dropped it altogether. But for $$\hat{r}$$ I'm unsure how to define it so that I can get a cross product. I was thinking that it should be equal to $$r cos \theta + r sin \theta$$ then take A X B = ABcos$$\theta$$ but since cross product is not distributive I'm just unsure.
I'm very confused about how to approach the geometry of the situation and this is not the first time this has been an issue for me. Any help is appreciated.

 

[gneill]

You say that the charge is at a position a > R from the axis of the solenoid, so it is outside the solenoid where the magnetic field is zero. Won't your cross products involving B be zero magnitude too?

 

[Bhumble]

For the portion outside the solenoid they will but there is still an electric and magnetic field inside the solenoid. But the electric field is still present inside the solenoid since the problem states it is a non-conductor.

 

[gneill]

Ah. So the charge is expressing its field inside the solenoid where there is a B field. Got it.

 

[rwisz]

First, let's define the coordinate system for this problem. We can choose the z-axis to be along the axis of the solenoid, the x-axis to be along the line connecting the point charge to the solenoid (as given in the problem), and the y-axis to be perpendicular to both the x and z axes.

Next, we can find the electric field at any point in space using Coulomb's law. Since the solenoid is treated as a non-conductor, we don't need to worry about induced charges on its surface. The electric field at a point (x,y,z) due to the point charge q is given by:

\vec{E} = \frac{q}{4\pi\epsilon_0}\frac{\hat{r}}{r^2}

where r is the distance from the point charge to the point (x,y,z) and \hat{r} is the unit vector pointing from the point charge to the point (x,y,z).

To find the linear momentum density, we need to calculate the cross product \vec{\rho_{em}} = \epsilon_0 \vec{E}\times\vec{B}. Since the electric field is only in the x-direction (due to the translational invariance along the z-axis), we can simplify this to:

\vec{\rho_{em}} = \epsilon_0 E_x B_z \hat{y}

where E_x is the x-component of the electric field and B_z is the z-component of the magnetic field. We can then integrate this over the surface area of the solenoid to find the total linear momentum:

\vec{p_{em}} = \int \vec{\rho_{em}} dA = \int \epsilon_0 E_x B_z \hat{y} dA

To find the angular momentum, we need to calculate the cross product \vec{L_{em}} = \vec{r}\times\vec{p_{em}}. Since the linear momentum is in the y-direction and the position vector is in the x-direction, the cross product will be in the z-direction. However, since the solenoid is infinite, the angular momentum will be zero, since the position vector will always be parallel to the linear momentum vector.

\vec{L_{em}} = \vec{r}\times\vec{p_{em}} = (x\hat{x}+y