
#1
Feb1609, 03:56 PM

P: 1,431

A cylindrical pipe of radius a and length L is filled with mercury of electrical conductivity [latex]\zeta[/latex]. 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 [latex]\mathbf{B}[/latex] at radius r from the axis. I said that [latex]\mathbf{J}=\zeta (\mathbf{E} + \mathbf{v} \wedge \mathbf{B})=\zeta \mathbf{E}[/latex] as [latex]\mathbf{v}=0[/latex] as mercury is stationary. Then I took an Amperian loop around the cylinder of radius r. Ampere's Law tells us that [latex]\oint_C \mathbf{B} \cdot \mathbf{dr}=\mu_0 \int_S \mathbf{J} \cdot \mathbf{dA}[/latex] Then, [latex]\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}[/latex] Firstly, how does this look up to this point? If ok, how do I find B from that last bit? cheers! 



#2
Feb1609, 08:44 PM

PF Gold
P: 363

[latex]\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}[/latex] 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 [latex] \oint_C dl [/latex] 



#3
Feb1609, 10:20 PM

P: 134





#4
Feb1709, 04:33 AM

P: 1,431

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 



#5
Feb1709, 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. 



#6
Feb1709, 08:38 AM

P: 1,431

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?




#7
Feb1709, 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.




#8
Feb1709, 09:10 AM

P: 1,431

i don't really follow. sorry.




#9
Feb1709, 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?




#10
Feb1709, 10:56 AM

P: 1,431

kl. cheers. would [latex]\oint_C dl = 2 \pi r [/latex] or [latex]2 \pi a[/latex]
also the surface integral  owuld it be [latex]\pi a^2[/latex] or [latex]\pi r^2[/latex] im gueesing they both involve r's as they're integrals round the amperian loop? 



#11
Feb1709, 11:07 AM

P: 1,431

i get [latex]\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[/latex]
so then [latex]\mathbf{B}(r)=\frac{1}{2} \mu_0 \zeta \mathbf{E} r \mathbf{\hat{\phi}}[/latex] how does that look? using that i need to show that the magnetic field, when acting on an infinitesimal current element [latex]\mathbf{J}dV[/latex] produces a radial force [latex]d \mathbf{F}=f \mathbf{\hat{r}} dV[/latex] 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??? 



#12
Feb1709, 12:30 PM

P: 134





#13
Feb1709, 12:41 PM

P: 1,431

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? 



#14
Feb1709, 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?




#15
Feb1709, 01:14 PM

P: 1,431

wouldn't this just be the same but set a=r as the amperian loop is now inside?




#16
Feb1709, 01:20 PM

P: 134

No, because r is not equal to a, it is in fact less than a.




#17
Mar2709, 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 crosssectional area of the pipe (a circle of radius "r" which is the precise location you'd like to know the Bfield 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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