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Magnetic field above rotating disk

  • #1

Homework Statement



What is the magnetic field a distance z above a rotating disk (ang.velocity w) with surface charge density inversely proportional to the square of the distance from the centre of the disk.

Homework Equations



K = v*surface charge density
B = (constant)*integral of (K x rhat / r^2 )da

The Attempt at a Solution



I just wanna know if i'm setting up this question right (basically, if i'm setting up the part in the integral correct, which is why i ommited the constant)

charge desntiy = c/s^2 where c is a constant (since it is inversely proportional to square of distance from centre of disk)
so K = v*charge density = ws*c/s^2 = wc/s
so K cross rhat = Ksin(alpha) where alpha is the angle between k and r
but in this case, the angle will always be 90 degrees since k and r are in perpendicular planes
so i need to inegrate (K/r^2)da
but all the horizontal components will cancel so i need to take only the vertical components, so i multiply by cos theta where theta is the angle r makes with the disk

so i have :
integral of (Kcostheta/r^2) da
costheta = s/(s^2+z^2)^0.5
K = wc/s
r^2 = s^2 + z^2

so i get
ingeral of wc/s * s/(s^2+z^2)^3/2 * da
da = sdsdtheta
so I have to integrate (wcs/(s^2+z^2)^3/2)dsdtheta from s=0 to s=R and theta= 0 to theta=2pi

is this correct?
 

Answers and Replies

  • #2
Also, what about if the surface charge density was proportional to the distance from the centre instead?
then I would get an integral of the type s^4/(s^2+z^2)^3/2

How in the world would i do this integral? The one above is much easier but I cant figure out a way to do this one
 

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