Vector Potential: Finding the Vector Potential of a Long Wire

[germx]

Hello all. I am hoping someone could lend a hand. I need to determine the vector potential of a long wire. I am given the B-field = ($$\mu$$I)/(2$$\pi$$r) and I know at some point I will need to use $$\nabla$$ X A. Thanks in advance.

 

[Phrak]

What's the equation of $$\bar{B}$$ in terms of $$\bar{A}$$?

The magnetic field runs in rings around the conductor, so it only an azimuthal component.
This tells you that the A field generating B will be limited to radial and axial components.

 

[astrokreng]

Hello there,

I am glad to assist you with finding the vector potential of a long wire. The vector potential, denoted by A, is a mathematical concept used to describe the magnetic field in terms of a vector quantity. In this case, the vector potential can be calculated using the formula A = (\muI)/(2\pi) \ln(r), where \mu is the permeability of the medium, I is the current flowing through the wire, and r is the distance from the wire.

To find the vector potential, we can use the fact that the magnetic field, denoted by B, is equal to \nabla X A, where \nabla is the gradient operator. Therefore, we can rewrite the formula for the vector potential as A = \frac{1}{\mu} \nabla X B.

In your case, you are given the B-field = (\muI)/(2\pi r), which is a function of r. We can use this information to calculate the vector potential by taking the cross product of the gradient operator and the B-field. This would result in A = \frac{1}{\mu} \left( \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} \right) \hat{x} + \frac{1}{\mu} \left( \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x} \right) \hat{y} + \frac{1}{\mu} \left( \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} \right) \hat{z}, where \hat{x}, \hat{y}, and \hat{z} are the unit vectors in the x, y, and z directions respectively.

I hope this helps you in finding the vector potential of the long wire. Let me know if you have any further questions. Good luck!