Vector Potential of Magnetized Cylinder

In summary, the conversation discusses finding the vector potential and magnetic field of a cylinder with a specified magnetization distribution. It is mentioned that the setup for finding the vector potential inside the cylinder is correct, but the contribution from surface currents needs to be included. It is also recommended to use cylindrical coordinates for simplification.
  • #1
stunner5000pt
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2

Homework Statement


Find the vector potential and magnetic field everyehere of a cylinder of radius R and length L which carries a magnetization [itex] \vec{M} = ks^2 \hat{\phi} [/itex] where k is a constant and s is the distance from the axis of the cylinder.

Homework Equations


[tex] A(r) = \frac{\mu_{0}}{4\pi}\int \frac{J_{b}(s')}{s'} d\tau' +\frac{mu_{0}}{4\pi} \int\frac{K_{b}(s')}{s'} da'[/tex]

[tex] B = \nabla \times A [/tex]

The Attempt at a Solution


Ok so let's consider the inside part
s<R
the surface charge is zero inside the cylinder
so
[tex] A(r) = \frac{\mu_{0}}{4\pi} \int\frac{J_{b}(s')}{s'} d\tau' [/tex]

[tex] J_{b}(s) = \nabla \times M = 3ks \hat{z} [/tex]

so [tex] A(r) = \frac{\mu_{0}}{4\pi} \int\frac{J_{b}(s')}{s'} d\tau' [/tex]

[tex] A(r) = \frac{\mu_{0}}{4\pi} \hat{z}\int\frac{3ks'}{s'} s'ds'd\phi' dz' [/tex]

is the setup right??

thanks for your help and advice!
 
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  • #2

Thank you for your post. Your setup for finding the vector potential inside the cylinder is correct. However, you have not taken into account the contribution from the surface current at the boundary of the cylinder. This can be included by adding the surface integral term in your expression for the vector potential, as shown in the homework equations. Additionally, to find the magnetic field, you will need to take the curl of the vector potential. I would recommend using cylindrical coordinates to simplify the integrals in your calculations.

I hope this helps. Good luck with your homework!

Scientist at [Your Institution]
 

1. What is the vector potential of a magnetized cylinder?

The vector potential of a magnetized cylinder is a mathematical concept used to describe the magnetic field produced by a cylindrical magnet. It is a vector quantity that has both magnitude and direction, and it is related to the magnetic dipole moment of the cylinder.

2. How is the vector potential of a magnetized cylinder calculated?

The vector potential of a magnetized cylinder can be calculated using the Biot-Savart law, which takes into account the current flowing through the cylinder and the distance from the cylinder. It can also be calculated using the curl of the magnetic field, as the vector potential is equal to the negative of the curl of the magnetic field.

3. What are the applications of the vector potential of a magnetized cylinder?

The vector potential of a magnetized cylinder has various applications in physics and engineering. It is used to calculate the magnetic field produced by a cylindrical magnet in a given space, and it is also used in the study of electromagnetic radiation and induction phenomena.

4. How does the vector potential of a magnetized cylinder differ from the scalar potential?

The vector potential of a magnetized cylinder is a vector quantity, meaning it has both magnitude and direction. In contrast, the scalar potential is a scalar quantity, meaning it only has magnitude. The vector potential is related to the magnetic field, while the scalar potential is related to the electric field.

5. Can the vector potential of a magnetized cylinder be measured?

While the vector potential itself cannot be measured directly, its effects can be observed and measured. For example, the magnetic field produced by the cylinder can be measured, and the vector potential can be calculated from the magnetic field using mathematical equations.

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