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Magnetic field of rotating cylinder 
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#1
Dec907, 11:32 AM

P: 219

1. The problem statement, all variables and given/known data
The problem is to find the magnetic field within a rotating cylinder (infinitely long) that has on its surface a given surface charge density p. I made a picture of the problem to illustrate this. The only hint given: "the magnetic field outside the cylinder is zero. 2. Relevant equations Sorry, I cannot use Latex, but: integral(B,ds) = u0*integral(J,da) where B is the magnetic field, ds the vector line element, J the current density, da the vector area element. 3. The attempt at a solution Well, to be cleary, I can solve this problem, but I do not understand a very basic thing. First I argued by symmetry, that the magnetic field can only be in direction of the angular frequency vector w. Then I took a curve c, illustrated in my picture and now I want to apply the integral law I have written above to this curve c. I know the left hand side of the integral, which is simply (given that the curve has lengt L): B*L, Now I want to calculate the right hand side. Doing so, I first need J. logically: J=(p/(2*pi*R))*(w*R) where (w*R) = v is simply the velocity. Up to here all is quite easy, but now I have some questions: i) Is it correct, that the J vector is the tangential vector to the outer surface of the cylinder perpendicular to the angular frequency vector w? ii) The integral law states that I have to integrate J over the area enclosed by the curve c. But this doesn't seems to be correct. I mean, the area within the cylinder and enclosed by the curve c would then be (Rr1)*L and outside undefined. So over which area do I have to integrate and WHY? 


#2
Dec907, 02:12 PM

HW Helper
PF Gold
P: 1,197

Is it a hollow cylinder? Also, do you have a surface current flowing only on the outside of the cylinder, or is it a volume current?



#3
Dec907, 03:12 PM

P: 219

Well, I think we can assume the cylinder to be a perfect conductor carrying only a surface charge density at its outside. So inside there will not be any current.



#4
Dec1007, 08:20 PM

HW Helper
PF Gold
P: 1,197

Magnetic field of rotating cylinder
So, you'll only need to calculate the net current enclosed by the curve C you drew. If [tex]\sigma[/tex] is the surface charge density, then the surface current will be [tex]K=\sigma v = \sigma \omega R[/tex]. From this, you can find the current enclosed by the loop and the magnetic field inside.



#5
Dec1107, 05:51 AM

P: 219

Ah, so you mean I have to integrate the surface current density K (unit A/m) over the line my curve intersects the cylinder? So I get K*length(curve) = K*L.



#7
Dec1207, 05:17 PM

P: 219

Ah, ok, then I understand it now. I think the first thing I were trying to do was integrating an object, namely the current density J, which is simply everywhere zero except at the boundaries of the cylinder over an area which intersects the current density vector J perpendicular. This cannot really work, because then J would be of zero measure for the integral of J over the area which is enclosed by the curve I have drawn (because with respect to the integral J*d(area), J is just defined on a line, namely the intersection of my rectangular area with the cylinder and zero elsewhere).
< This seems to be written in a strange manner, but well, this was just what confused me. Thank you for help. 


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