Magnetic field of rotating sphere

  • #1
912
19
Hi

I have to find the magnetic field at the center of a rotating sphere
which has unifrom charge density [itex]\rho[/itex] on it. The sphere
is rotating with angular velocity [itex]\omega [/itex] . The answer given
in the solution manual is

[tex] B= \frac{1}{3} \mu_o \rho \omega R^2 [/tex]

I didn't get this value. Let me present my work here. I first found the
magnetic field at the center of a disk, which has uniform charge density
[itex]\sigma[/itex] on it. After this, I extend this to the solid sphere as
collection of disks. Now the magnetic field at the center of a ring of radius
r , is

[tex] B = \frac{\mu_o I r^2}{2(x^2+r^2)^{\frac{3}{2}}} [/tex]

where x is the distance of some point on the axis of the ring from
the center of the ring. If the ring is rotating with angular velocity
[itex]\omega[/itex], then the current , I , is

[tex] I = \frac{q ' \omega}{2 \pi} [/tex]

Now, if this ring is the part of a disk, then we have

[tex] q ' = \sigma (2\pi r) dr [/tex]

plugging everything we have, the magnetic field due to a ring, which is
a part of the disk, at a distance x (from the center) on the axis of the ring is

[tex] dB= \frac{\mu_o \omega \sigma}{2} \,\, \frac{r^3 \, dr}{\left(x^2+r^2 \right)^{\frac{3}{2}}} [/tex]

Now , we integrate this from r=0 to r=R , which is the radius of the
disk.

[tex]B= \frac{\mu_o \omega \sigma}{2}\,\, \int_0^R \frac{r^3 \, dr}{\left(x^2+r^2 \right)^{\frac{3}{2}}} [/tex]

after integrating , I got the magnetic field , B , of the disk , at a distance
x from the center of the disk, on its axis

[tex] B= \frac{\mu_o \omega \sigma}{2}\,\, \left[ \frac{2x^2+R^2}{(x^2+R^2)^{\frac{1}{2}}} - 2x \right ] [/tex]

Now , the sphere can be thought as a collection of disks. Lets consider the disk, at an angle
[itex]\theta [/itex] from the z-axis. So the radius of the disk would be [itex]R \sin \theta[/itex].
The thickness of the disk would be [itex]R \,d \theta \, \sin \theta [/itex] . I think I got the thickness
right. I am not sure about that. So the relationship between the surface charge density
on the disk and the volume charge density in the sphere is

[tex]\sigma = \rho \, R \,\, d\theta (\sin \theta) [/tex]

So , dimensionally this makes sense. So plugging these values in the expression for
the magnetic field due to disk, I get

[tex]dB=\frac{\mu_o \omega }{2} (\rho R d \theta \sin \theta) \left[R(1+\cos^2 \theta) - 2R \cos \theta \right] [/tex]

So we integrate this from [itex]\theta = 0 [/itex] to [itex]\theta = \pi [/itex] to cover the whole sphere.
And I get

[tex] B= \frac{4}{3} \mu_o \rho \omega R^2 [/tex]

But as I said earlier, the reas answer is given as

[tex] B=\frac{1}{3} \mu_o \rho \omega R^2 [/tex]

So where am I going wrong ?
 

Answers and Replies

  • #2
106
1
Your method seems fine until the end, when you integrate over theta. Instead of integrating over the whole sphere, try finding the magnetic field due to half the sphere, then simply doubling that.

That said, I'm not quite sure why that works while your method doesn't. My intuition would be that something funny is happening with the [tex]-2R\cos(\theta)[/tex] term in [tex]dB[/tex]. If you think about it conceptually, a disc at [tex]\theta = \pi/4[/tex] should produce a magnetic field of the same magnitude as a disc at [tex]\theta = 3\pi/4}[/tex]. In terms of the magnitude of the magnetic field, the entire sphere should be symmetric across the [tex]\theta = \pi/2[/tex] plane. However, the cosine term I mentioned above will flip signs as one crosses this plane, causing the magnitude of the field given by your equation to be asymmetric. I'll leave it up to someone smarter and less sleep-deprived than me to explain why this issue exists.
 
  • #3
Chronos
Science Advisor
Gold Member
11,408
741
If the charge is uniformly distributed ... worth some thought.
 
  • #4
912
19
Leveret,

thanks for the input but I am not convinced by what you're trying to say.
What funny thing is happening ?

Chronos, what is worth some thought ?
 
  • #5
I like Serena
Homework Helper
6,577
176
As I understand electrostatics, if you electrically charge a sphere, whether it is massive or not, all electrical charge will be located at the outside of the sphere.
I remember that you can proof this.

So I believe you shouldn't use disks, but rings.
Each ring would have a radius r = √(R²-z²) and a thickness Rdθ.
Each ring would then have a charge 2πρrRdθ.

Next you would integrate z from -R to +R.

Effectively the charge would be further away from the center, resulting in a lower value for B.
 
  • #6
912
19
As I understand electrostatics, if you electrically charge a sphere, whether it is massive or not, all electrical charge will be located at the outside of the sphere.
Well, thats true for a conducting sphere. I don't think we have a conducting sphere. The problem explicitly mentions that charge is uniformly distributed over all volume.
 

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