Magnetic Moment of a spinning shell

So, in summary, your solution is incorrect because it does not consider the changing angular velocity of the sphere.
  • #1
Ataman
18
0
The problem reads:

Find the magnetic moment of a spinning shell of charge Q, radius R, and angular velocity [tex]\vec{\omega}[/tex].

My solution:

I split the sphere into infinitely small loops of current (from top to bottom) and add them up. I set my origin on the center of the sphere and integrate with respect to y from 0 to R times two to get the entire sphere.

The start of the problem:

[tex]\vec{\mu} \,= A\vec{I}[/tex]
[tex]d\vec{\mu} \,= A(d\vec{I})[/tex]

Solving for [tex]d\vec{I}[/tex]:

[tex]d\vec{I} = \frac{dQ}{t}[/tex]
[tex]d\vec{I} = \frac{\sigma dA}{t}[/tex]
[tex]d\vec{I} = \frac{\sigma dA}{(\frac{2\pi}{\omega})}[/tex]
[tex]d\vec{I} = \frac{\sigma \omega dA}{2\pi}[/tex]

Plugging [tex]d\vec{I}[/tex] back:

[tex]d\vec{\mu} \,= \frac{\sigma \omega A dA}{2\pi}[/tex]

Changing A and dA, so I could integrate in terms of dy:

[tex]A \,= \pi r^2[/tex]
[tex]A \,= \pi (R^2-y^2)[/tex]
[tex]dA \,= -2 \pi y dy[/tex]

Then setting up the integral:
[tex]d\vec{\mu} \,= - \frac{2 \sigma \omega \pi^2}{2\pi} y (R^2-y^2) dy[/tex]
[tex]d\vec{\mu} \,= - \sigma \omega \pi y (R^2-y^2) dy[/tex]
[tex]\vec{\mu} \,= - 2 \sigma \omega \pi \int_{0}^{R} y (R^2-y^2) dy[/tex]

Without sigma:
[tex]\vec{\mu} \,= - \frac {Q \omega}{2R^2} \int_{0}^{R} y (R^2-y^2) dy[/tex]

This is incorrect, and I have a suspicion it has to do with the "dA" part.

If you don't mind his handwriting, http://nebula.deanza.fhda.edu/physics/Newton/4B/MagneticMomentSphere.jpg is the correct way to do the problem. I am rather uncomfortable integrating over angles, so I opted for a different way.

I would like to know why my way does not work.

-Ataman
 
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  • #2
The problem with your solution is that you are not taking into account the fact that the angular velocity \vec{\omega} is changing over the surface of the sphere. This means that the amount of current in each loop (and thus the magnetic moment) is not constant. In order to calculate the magnetic moment, you need to take into account the angular velocity at each point on the surface of the sphere. The solution given in the link you provided does this by integrating the magnetic moment over the entire surface of the sphere.
 

Related to Magnetic Moment of a spinning shell

1. What is the definition of magnetic moment of a spinning shell?

The magnetic moment of a spinning shell refers to the measure of the strength and direction of the magnetic field produced by a shell or ring of electrically charged particles that are rotating or spinning.

2. How is the magnetic moment calculated for a spinning shell?

The magnetic moment of a spinning shell can be calculated by multiplying the current flowing through the shell by the area that is enclosed by the shell and the axis of rotation. It can also be calculated by multiplying the angular momentum of the spinning shell by the gyromagnetic ratio.

3. What factors affect the magnetic moment of a spinning shell?

The magnetic moment of a spinning shell is influenced by various factors such as the strength of the current flowing through the shell, the radius of the shell, the speed of rotation, and the type of material that makes up the shell.

4. How does the magnetic moment of a spinning shell relate to the Earth's magnetic field?

The Earth's magnetic field can be thought of as a magnetic dipole, and the magnetic moment of a spinning shell is also a measure of a magnetic dipole. Therefore, the magnetic moment of a spinning shell can be used to understand and model the Earth's magnetic field.

5. What are some real-world applications of the magnetic moment of a spinning shell?

The magnetic moment of a spinning shell has many practical applications, such as in magnetic resonance imaging (MRI) machines, where it is used to create a strong and stable magnetic field. It is also used in particle accelerators and in the study of magnetic materials and their properties.

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