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Magnetostatics

by latentcorpse
Tags: magnetostatics
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latentcorpse
#1
Feb16-09, 03:56 PM
P: 1,441
A cylindrical pipe of radius a and length L is filled with mercury of electrical conductivity [itex]\zeta[/itex]. A potential difference V acts across the two ends of the pipe, creating an electric current through the mercury (which remains stationary).

(i)Find the current density, assumed uniform, within the mercury. Using Ampere's Law, find the magnetic field [itex]\mathbf{B}[/itex] at radius r from the axis.

I said that [itex]\mathbf{J}=\zeta (\mathbf{E} + \mathbf{v} \wedge \mathbf{B})=\zeta \mathbf{E}[/itex] as [itex]\mathbf{v}=0[/itex] as mercury is stationary.

Then I took an Amperian loop around the cylinder of radius r.
Ampere's Law tells us that [itex]\oint_C \mathbf{B} \cdot \mathbf{dr}=\mu_0 \int_S \mathbf{J} \cdot \mathbf{dA}[/itex]
Then,
[itex]\oint_C \mathbf{B} \cdot \mathbf{dr} = \mu_0 |\mathbf{J}| \int_S dA \Rightarrow \mu_0 \zeta |\mathbf{E}| \pi r^2 = \oint_C \mathbf{B} \cdot \mathbf{dr}[/itex]

Firstly, how does this look up to this point?
If ok, how do I find B from that last bit?

cheers!
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AEM
#2
Feb16-09, 08:44 PM
PF Gold
P: 362
Quote Quote by latentcorpse View Post
A cylindrical pipe of radius a and length L is filled with mercury of electrical conductivity [itex]\zeta[/itex]. A potential difference V acts across the two ends of the pipe, creating an electric current through the mercury (which remains stationary).

(i)Find the current density, assumed uniform, within the mercury. Using Ampere's Law, find the magnetic field [itex]\mathbf{B}[/itex] at radius r from the axis.

I said that [itex]\mathbf{J}=\zeta (\mathbf{E} + \mathbf{v} \wedge \mathbf{B})=\zeta \mathbf{E}[/itex] as [itex]\mathbf{v}=0[/itex] as mercury is stationary.

Then I took an Amperian loop around the cylinder of radius r.
Ampere's Law tells us that [itex]\oint_C \mathbf{B} \cdot \mathbf{dr}=\mu_0 \int_S \mathbf{J} \cdot \mathbf{dA}[/itex]
Then,
[itex]\oint_C \mathbf{B} \cdot \mathbf{dr} = \mu_0 |\mathbf{J}| \int_S dA \Rightarrow \mu_0 \zeta |\mathbf{E}| \pi r^2 = \oint_C \mathbf{B} \cdot \mathbf{dr}[/itex]

Firstly, how does this look up to this point?
If ok, how do I find B from that last bit?

cheers!
Since the letter 'r' refers to the radius within the pipe, I think you would be better off with a different variable of integration in the this last equation, say

[itex]\oint_C \mathbf{B} \cdot \mathbf{dl} = \mu_0 |\mathbf{J}| \int_S dA \Rightarrow \mu_0 \zeta |\mathbf{E}| \pi r^2 = \oint_C \mathbf{B} \cdot \mathbf{dl}[/itex]

Then, because you are integrating along the circumference of a circle, B and your line element dl are parallel so the cosine part of the dot product is 1 and your integral is very easy (oh, B is a constant on a circle of constant radius in this situation so it comes out of the integral.) What is

[itex] \oint_C dl [/itex]
xboy
#3
Feb16-09, 10:20 PM
P: 134
Quote Quote by latentcorpse View Post

I said that [itex]\mathbf{J}=\zeta (\mathbf{E} + \mathbf{v} \wedge \mathbf{B})=\zeta \mathbf{E}[/itex] as [itex]\mathbf{v}=0[/itex] as mercury is stationary.
There's some problem with this logic of putting v = 0 . If there's a current, it means charged particles are moving and have some velocity. But is there a magnetic field acting on this current?

latentcorpse
#4
Feb17-09, 04:33 AM
P: 1,441
Magnetostatics

well no there is no field so B=0, but surely the movement of charged particles will induce a magnetic field?
also how do we know B is constaant on a circle of constant radius
xboy
#5
Feb17-09, 07:41 AM
P: 134
Yes, but that field won't act on the source itself.
We know B is constant from the cylindrical symmetry of the situation. No matter where you stand on the circle, the situation looks just the same. You can see the same amount of current flowing through the wire, and at the same distance. So it follows that you get the same field at all points.
latentcorpse
#6
Feb17-09, 08:38 AM
P: 1,441
ok i understand the azimuthal symmetry bit but why would the field not act on the source is it because B is radially outward and v is perpendicular to that so wedge product would be 0?
xboy
#7
Feb17-09, 08:49 AM
P: 134
No, it can't produce a force on itself for the same reason you can't pull yourself up on a bucket - the source can't move itself w/o some external force appearing.
latentcorpse
#8
Feb17-09, 09:10 AM
P: 1,441
i don't really follow. sorry.
xboy
#9
Feb17-09, 09:21 AM
P: 134
Think of the electrostatic case. An isolated charged body creates a field. Now if the field acts on this charged body itself, the isolated charged body begins to move. This violates conservation of momentum. Does this make things clearer?
latentcorpse
#10
Feb17-09, 10:56 AM
P: 1,441
kl. cheers. would [itex]\oint_C dl = 2 \pi r [/itex] or [itex]2 \pi a[/itex]
also the surface integral - owuld it be [itex]\pi a^2[/itex] or [itex]\pi r^2[/itex]

im gueesing they both involve r's as they're integrals round the amperian loop?
latentcorpse
#11
Feb17-09, 11:07 AM
P: 1,441
i get [itex]|\mathbf{B}|=\frac{\mu_0 \zeta |\mathbf{E}| \int_S dA}{\oint_C dl}=\frac{\mu_0 \zeta |\mathbf{E}| \pi r^2}{2 \pi r} = \frac{1}{2} \mu_0 \zeta |\mathbf{E}|r[/itex]
so then

[itex]\mathbf{B}(r)=\frac{1}{2} \mu_0 \zeta |\mathbf{E}| r \mathbf{\hat{\phi}}[/itex]

how does that look?

using that i need to show that the magnetic field, when acting on an infinitesimal current element [itex]\mathbf{J}dV[/itex] produces a radial force [itex]d \mathbf{F}=f \mathbf{\hat{r}} dV[/itex] and determine f.
do i use the lorentz force formula for this part? as v is in z direction so cross product would give a radial force???
xboy
#12
Feb17-09, 12:30 PM
P: 134
Quote Quote by latentcorpse View Post
kl. cheers. would [itex]\oint_C dl = 2 \pi r [/itex] or [itex]2 \pi a[/itex]
also the surface integral - owuld it be [itex]\pi a^2[/itex] or [itex]\pi r^2[/itex]

im gueesing they both involve r's as they're integrals round the amperian loop?
While evaluating the surface integral keep in mind that there is no current outside the cylinder, so all contributions to the integral come from inside the cylinder.
latentcorpse
#13
Feb17-09, 12:41 PM
P: 1,441
so surface integral is pi a^2 and loop integral is 2 pi r (as we go all the way round our amperian loop)
kl

any advice for the next part?
xboy
#14
Feb17-09, 01:02 PM
P: 134
Actually the first part isn't done yet. What you found out here was the field at a distance r > a. what about r< a?
latentcorpse
#15
Feb17-09, 01:14 PM
P: 1,441
wouldn't this just be the same but set a=r as the amperian loop is now inside?
xboy
#16
Feb17-09, 01:20 PM
P: 134
No, because r is not equal to a, it is in fact less than a.
physicsboy
#17
Mar27-09, 09:01 AM
P: 6
The voltage on the plates creates a uniform electric field inside. E=vL. You can use the idea that the current density is proportional to the electric field. J=\sigma E where \sigma is the conductivity. Use the Maxwell equation curl(B) = (4\pi)/c J and integrate it over a cross-sectional area of the pipe (a circle of radius "r" which is the precise location you'd like to know the B-field at). Ultimately you get \int(B dot dr) = (4 \pi)/c \int(\sigma E dot dA). Assume B is in the direction of the loop (the "phi" direction if you like) and E and dA are in the same direction. 2 \pi r B = (4 \pi)/c \sigma vL \pi a^2. Thus B = (2 \pi \sigma)/c vLa^2/r. B varies like 1/r.


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