Finding Angular & Linear Momentum of EM Fields

In summary, to find the linear and angular momentum for a point charge q at a distance a>R from the axis of an infinite solenoid, we use the equations \vec{E}q=q/4\pi\epsilon0(1/\vec{r}2)=q/4\pi\epsilon0(\vec{r}/r3) and \vec{B}sol=μ0nI\hat{z}, along with the formula Pem=ε0(\vec{E}\times\vec{B}) and the integral equations Pem=∫pem d\tau and Lem=∫lem d\tau. By converting the integral to cylindrical coordinates and using some tricky calculus, we can
  • #1
FlatLander
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0

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 you don't need to worry about induced charges on its surface.)[Answer: Pem0qnIR2/2a; Lem=0]

Homework Equations


[itex]\vec{E}[/itex]q=q/4[itex]\pi[/itex][itex]\epsilon[/itex]0(1/[itex]\vec{r}[/itex]2)=q/4[itex]\pi[/itex][itex]\epsilon[/itex]0([itex]\vec{r}[/itex]/r3)
[itex]\vec{B}[/itex]sol0nI[itex]\hat{z}[/itex]
pem0([itex]\vec{E}[/itex][itex]\times[/itex][itex]\vec{B}[/itex])
lem=r[itex]\times[/itex]pem
Pem=∫pem d[itex]\tau[/itex]
Lem=∫lem d[itex]\tau[/itex]

The Attempt at a Solution


I kind of plugged and chugged, found r2=((x-a)2+y2+z2) and [itex]\vec{r}[/itex]=(x-a)[itex]\hat{x}[/itex]+y[itex]\hat{y}[/itex]+z[itex]\hat{z}[/itex]
Plugged in for that as well. However, I eventually got to the integrations in for the Pem and realized I don't know what my limits of integration are for the volume.
I know the z is from -∞ to ∞, but I have no clue for x and y. Here is what my final line looks like so far (with me already integrating over z):
Pem=[itex]\frac{-2\mu_{0}\epsilon_{0}qnI}{4\pi\epsilon_{0}}[/itex]∫[itex]\frac{(x-a)\hat{y}}{((x-a)^{2}+y^{2}}[/itex] dydx

Any help would be appreciated. Thanks.
 
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  • #2
I figured it out. I'm too lazy to type out the solution, and since I know this could potentially help someone in the future searching for it, I'll give you the basic steps:
1. Change the integral I had in the last line to cylindrical coordinates.
[From here it's just tricky calculus]
2. Do the phi integral first and break it up into two parts
3. Here's where it gets tricky. You need to define two new variables, say A and B, where A=s^2+a^2, B=-2as.
4. Do the integrals. DO NOT do them with a calculator or other computation engine. It WILL give you wrong answer. Use an integral table or do them by hand (good luck).
5. Finding Lem is trivial from this point on, you find lem, then do the integral. Repeat steps 1-4 with the integral you get from Lem.
 

FAQ: Finding Angular & Linear Momentum of EM Fields

What is angular momentum of an EM field?

Angular momentum of an EM field refers to the measure of the rotational motion of an electromagnetic field. It is a vector quantity that is defined as the cross product of the position vector and the linear momentum of the field.

How is angular momentum calculated for an EM field?

The angular momentum of an EM field can be calculated by multiplying the position vector of the field by its linear momentum and the sine of the angle between them. This can also be expressed as the product of the field's moment of inertia and its angular velocity.

What is linear momentum of an EM field?

Linear momentum of an EM field refers to the measure of the straight-line motion of the field. It is a vector quantity that is defined as the product of the field's mass and its velocity.

How is linear momentum calculated for an EM field?

The linear momentum of an EM field can be calculated by multiplying the field's mass by its velocity. This can also be expressed as the product of the field's density, volume, and its velocity.

What is the relationship between angular and linear momentum of an EM field?

The angular momentum and linear momentum of an EM field are related through the cross product formula. This means that any change in the angular momentum of the field will result in a corresponding change in its linear momentum, and vice versa.

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