# Magnetic moment of a solid, uniformly charged ball

## 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:

$\textbf{μ} = \frac{5q}{6mc}\textbf{L}$

## Homework Equations

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

## The Attempt at a Solution

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

$dμ = dI \cdot dA$

where dA is the area enclosed by the segment:

$dA = 2π r dr$

The current dI on each segment is:

$dI = \frac{dq}{T} = \frac{ω}{2π} dq$

The charge on each segment is:

$dq = σ da$

Where da is the surface area of the segment:

$da = r^2\sinθ dθ d\phi$

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

We now have an expression for dI:

$dI = \frac{ωqr^2\sinθ}{8π^2R^2} dθ d\phi$

Now we can calculate the moment of each segment:

$dμ = \frac{ωqr^2\sinθ}{8π^2R^2} 2πr dr d\phi dθ$

Integrating this over the whole sphere gives

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

$μ = \frac{wqR^2}{8}$

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

$μ = \frac{wqR^2}{3}$

Where am I going wrong?

Last edited:

TSny
Homework Helper
Gold Member
Split sphere into thin segments, each one has magnetic moment:

$dμ = dI \cdot dA$

where dA is the area enclosed by the segment:

$dA = 2π r dr$

Is that the correct expression for the area enclosed? Can you describe the area that you are talking about? Did you make a sketch?

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:

$A = πr^2\sin^2θ$

This may be where I am going wrong but I am struggling to see why the radius is rsinθ

Last edited:
TSny
Homework Helper
Gold Member
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

• Area Spherical Strip.png
2.3 KB · Views: 566
I see it now, thank you!

The radius of the segment is just simple trigonometry,

$r = R\sinθ$

So the area now becomes:

$A = πR^2 \sin^2θ$

TSny
Homework Helper
Gold Member
Good.

• 1 person