Magnetic moment of a solid, uniformly charged ball

  • Thread starter six7th
  • Start date
  • #1
14
0

Homework Statement


Show for a solid spherical ball of mass m rotating through its centre with a charge q uniformly distributed on the surface of the ball that the magnetic moment μ is related to the angular momentum L by the relation:

[itex]\textbf{μ} = \frac{5q}{6mc}\textbf{L}[/itex]


Homework Equations



[itex]μ = IA [/itex]
[itex]I = \frac{2}{5}MR^2[/itex]
[itex]σ = \frac{q}{4πR^2}[/itex]
[itex]\frac{1}{T} = \frac{ω}{2π}[/itex]


The Attempt at a Solution



Split sphere into thin segments, each one has magnetic moment:

[itex]dμ = dI \cdot dA [/itex]

where dA is the area enclosed by the segment:

[itex]dA = 2π r dr[/itex]

The current dI on each segment is:

[itex]dI = \frac{dq}{T} = \frac{ω}{2π} dq[/itex]

The charge on each segment is:

[itex]dq = σ da[/itex]

Where da is the surface area of the segment:

[itex]da = r^2\sinθ dθ d\phi[/itex]

This results in
[itex]dq = σ \cdot r^2\sinθ dθ d\phi = \frac{q}{4πR^2} r^2\sinθ dθ d\phi[/itex]

We now have an expression for dI:

[itex]dI = \frac{ωqr^2\sinθ}{8π^2R^2} dθ d\phi[/itex]

Now we can calculate the moment of each segment:

[itex]dμ = \frac{ωqr^2\sinθ}{8π^2R^2} 2πr dr d\phi dθ[/itex]

Integrating this over the whole sphere gives

[itex]μ = \frac{ωq}{4πR^2} \int^{R}_{0} r^3 dr \int^{π}_{0} \sinθ dθ \int^{2π}_{0} d\phi [/itex]

[itex]μ = \frac{wqR^2}{8}[/itex]

Obviously this is not correct, the answer I should be getting is

[itex]μ = \frac{wqR^2}{3}[/itex]

Where am I going wrong?
 
Last edited:

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,966
3,322
Split sphere into thin segments, each one has magnetic moment:

[itex]dμ = dI \cdot dA [/itex]

where dA is the area enclosed by the segment:

[itex]dA = 2π r dr[/itex]
Is that the correct expression for the area enclosed? Can you describe the area that you are talking about? Did you make a sketch?
 
  • #3
14
0
For a normal loop of wire, A would be the area inside the loop so in this case dA is the area bounded by the surface, da, of the thin segment. I have tried using the volume of the segments but the equation for magnetic moments requires an area, so I couldn't see how this would work.

Edit: I have seen on similar problems that the radius of each segment is rsinθ making the area enclosed:

[itex]A = πr^2\sin^2θ[/itex]

This may be where I am going wrong but I am struggling to see why the radius is rsinθ
 
Last edited:
  • #4
TSny
Homework Helper
Gold Member
12,966
3,322
The area enclosed by a circular strip on the sphere is shown in yellow in the figure. Can you express that area in terms of R and θ?
 

Attachments

  • #5
14
0
I see it now, thank you!

The radius of the segment is just simple trigonometry,

[itex]r = R\sinθ [/itex]

So the area now becomes:

[itex]A = πR^2 \sin^2θ [/itex]
 
  • #6
TSny
Homework Helper
Gold Member
12,966
3,322
Good.
 
  • Like
Likes 1 person

Related Threads on Magnetic moment of a solid, uniformly charged ball

Replies
7
Views
19K
Replies
4
Views
7K
Replies
2
Views
20K
Replies
6
Views
12K
Replies
4
Views
12K
Replies
1
Views
2K
Replies
1
Views
5K
Replies
4
Views
11K
Replies
4
Views
5K
Top