How to Prove the Magnetic Induction B from Vector Potential A Elegantly?

[Sagreda]

I'm having a hard time proving that if $$A=k \frac{m\times r}{(r\cdot r)^{3/2}}$$ (the vectorpotential of a magnetic dipole with moment $$m$$), then:

$$B=\nabla\times A=k\frac{3e_r(e_r\cdot m) -m}{(r\cdot r)^{3/2}}$$

without writing the whole thing in components, which becomes long, messy and ugly.
What's the elegant way of proving the magnetic induction $$B$$ in vectorial form from the vector potential $$A$$?

 

[Gerenuk]

I've attached a page from my vector notes. There you see the basic equations to treat these kind of problems. There first part I found in a book which one can check by components.

In the second part I derived some particular results from the general equations. Feel free to check them, as I haven't done that yet.

There you also find your case.

 

[tacman]

Vector calculus can definitely be intimidating, but once you understand the concepts, it becomes much easier to see the elegant way of proving relationships such as the one you mentioned. In this case, the key concept to understand is that the magnetic field, B, is defined as the curl of the vector potential, A. This means that B can be written as the cross product of the gradient operator and A:

B = ∇ x A

Using this definition, we can see that the components of B can be written as:

B_x = (∂A_z/∂y) - (∂A_y/∂z)
B_y = (∂A_x/∂z) - (∂A_z/∂x)
B_z = (∂A_y/∂x) - (∂A_x/∂y)

Now, let's focus on just one of these components, say B_x. We can rewrite it as:

B_x = (∂A_z/∂y) - (∂A_y/∂z)
= (∂/∂y)(k(m x r)/(r^2)^(3/2)) - (∂/∂z)(k(m x r)/(r^2)^(3/2))
= k(∂/∂y)((m x r)/(r^2)^(3/2)) - k(∂/∂z)((m x r)/(r^2)^(3/2))
= k(∂/∂y)(3e_r(m∙e_r) - m)/(r^2)^(3/2) - k(∂/∂z)(3e_r(m∙e_r) - m)/(r^2)^(3/2)
= k(3e_r(m∙∂e_r/∂y) - m∙∂/∂y(e_r))/(r^2)^(3/2) - k(3e_r(m∙∂e_r/∂z) - m∙∂/∂z(e_r))/(r^2)^(3/2)
= k(3e_r(m∙e_y) - m∙0)/(r^2)^(3/2) - k(3e_r(m∙e_z) - m