# The Curl of Magnetic Potential

• quantumfoam
In summary, the conversation discusses the calculation of the curl of a magnetic potential, A(r), in terms of its Cartesian components. The result is found to be (u/4π)([3(m•r)R/r^5]-[m/r^3]) after using the identity for the cross product of two vectors and simplifying terms. It is also mentioned that using the spherical version of the del operator may make the calculation easier.
quantumfoam
Hi guys! I recently saw on Wiki that given a magnetic potential A(r)=(u/4π)(mXR/r^3) ,( where u is the permeability of free space, m is the magnetic dipole moment of a magnetic field, R is the position vector, and r is the distance from the magnetic field ) upon taking the curl of the magnetic potential A(r), one obtains (u/4π)([3(m•r)R/r^5]-[m/r^3]). Can someone show me how they ended up with this result? I would like a step by step process please.(: the wiki article is called Magnetic Moment under the section called Effects of an External Magnetic Field on a Magnetic Moment.

quantumfoam said:
Hi guys! I recently saw on Wiki that given a magnetic potential A(r)=(u/4π)(mXR/r^3) ,( where u is the permeability of free space, m is the magnetic dipole moment of a magnetic field, R is the position vector, and r is the distance from the magnetic field ) upon taking the curl of the magnetic potential A(r), one obtains (u/4π)([3(m•r)R/r^5]-[m/r^3]). Can someone show me how they ended up with this result?

I would like a step by step process please.(: the wiki article is called Magnetic Moment under the section called Effects of an External Magnetic Field on a Magnetic Moment.
That's not how we do things in the Homework section of PF. We're not even allowed to give such help. But we can try to help you work your way through such a problem.

One way to get the result, is to write A(r) in terms of it's Cartesian components, then take the curl. I suppose you could also take the curl directly in spherical coordinates.

I'm sorry! Well I thought about doing it in spherical coordinates but I was not sure if I should use the whole three partials in the del operator.

quantumfoam said:
I'm sorry! Well I thought about doing it in spherical coordinates but I was not sure if I should use the whole three partials in the del operator.
Yes, use all three partials.

quantumfoam said:
I'm sorry! Well I thought about doing it in spherical coordinates but I was not sure if I should use the whole three partials in the del operator.

I don't think there is any advantage is using spherical coordinates. Try using the identity:

$$\nabla \times (a \times b)=(b \cdot \nabla) a-b(\nabla \cdot a) -(a \cdot \nabla) b+a(\nabla \cdot b)$$

If you set a to be m and b to be ##\frac{\vec r}{r^3}##, the first two terms vanish because m is a constant. The last one vanishes because ##\frac{\vec r}{r^3}## is divergence free. So you only have to puzzle out what the third term means.

Thank you very much! (:

Is there any other identity I could use for (a(div))b? I did the calculation but I think I did it wrong. I'm missing the second term in the curl of the magnetic potential. The (m/r^3) term

quantumfoam said:
Is there any other identity I could use for (a(div))b? I did the calculation but I think I did it wrong. I'm missing the second term in the curl of the magnetic potential. The (m/r^3) term

Show what you did. You'll need to use the quotient rule on the dervivatives ##\frac{\vec r}{r^3}##.

Well, at first I used the identity (a(div))b= (div(a))b-b(div(a)). Was that a correct move?

Oh! I get it! Nevermind! I didn't think to carry out the quotient rule. That makes a lot more sense! Thank you!

So I should calculate using the radial term of the spherical version of the del operator?

That's the only way I see to take the derivative of R/r^3.

quantumfoam said:
That's the only way I see to take the derivative of R/r^3.

R/r^3 is just (x,y,z)/(x^2+y^2+z^2)^(3/2). It worked out pretty ok for me in cartesian coordinates.

Would it be a little easier with spherical coordinates?

quantumfoam said:
Would it be a little easier with spherical coordinates?

Maybe. I didn't try it that way. I just know it's not hard in cartesian.

I will try it! Thank you!

## What is the curl of magnetic potential?

The curl of magnetic potential is a mathematical concept used in electromagnetism to describe the behavior of magnetic fields. It is represented by the symbol ∇ × A and measures the tendency of a magnetic field to circulate around a given point or axis.

## How is the curl of magnetic potential calculated?

The curl of magnetic potential is calculated using the vector calculus operator known as the gradient. This involves taking the partial derivatives of the magnetic potential function with respect to the three spatial coordinates (x, y, and z).

## What does a high curl of magnetic potential indicate?

A high curl of magnetic potential indicates a strong magnetic field that is concentrated around a specific point or axis. This can be seen in areas with strong magnetic forces, such as near a magnet or in the vicinity of an electric current.

## What is the significance of the curl of magnetic potential in physics?

The curl of magnetic potential is a fundamental concept in electromagnetism and has important applications in physics, particularly in understanding the behavior of magnetic fields. It is used to calculate the magnetic field strength and direction at any point in space and is essential in many practical applications, such as in the design of electronic devices and in studying celestial bodies.

## Is the curl of magnetic potential the same as the curl of electric potential?

No, the curl of magnetic potential and the curl of electric potential are two different mathematical concepts. The curl of electric potential, represented by the symbol ∇ × φ, describes the behavior of electric fields and is used in the study of electrostatics. However, they are related through Maxwell's equations, which describe the relationship between electric and magnetic fields.

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