# Magnetic field of vector potential

## Homework Statement:

Using the vector potential A, show that the Cartesian representation of the magnetic induction field associated with a magnetic moment oriented along the Cartesian z-axis is B

## Relevant Equations:

$$\vec{B} = \nabla \times \vec{A}$$
$$\vec{A} = \frac{\mu}{4\pi}\frac{m_z}{r^3} (-y,x,0)$$
$$\vec{B} = \frac{\mu}{4\pi}\frac{m_z}{r^5} (3xz, 3yz, 3z^2-r^2)$$
$$\frac{\partial}{\partial x} \frac{1}{r^3} = -\frac{3x}{r^5}$$
So I was able to do out the curl in the i and j direction and got 3xz/r5 and 3yz/r5 as expected. However, when I do out the last curl, I do not get 3z2-3r2. I get the following
$$\frac{\partial}{\partial x} \frac{x}{(x^2+y^2+z^2)^\frac{3}{2}} = \frac{-2x^2+y^2+z^2}{(x^2+y^2+z^2)^\frac{5}{2}}$$
$$\frac{\partial}{\partial y} \frac{-y}{(x^2+y^2+z^2)^\frac{3}{2}} = \frac{-2y^2+x^2+z^2}{(x^2+y^2+z^2)^\frac{5}{2}}$$
which when added together gives me
$$\frac{-x^2-y^2+2z^2}{(x^2+y^2+z^2)^\frac{5}{2}}$$.

I can't see where I've gone wrong with this differentiation. I've tried it out on symbolab and get the same result.

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TSny
Homework Helper
Gold Member
Could it be that your answer is equivalent to what you are trying to get?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Expand ##r^2## in terms of x y and z ...

Good God, I'm a moron. Thanks to you both. I got it out there

EDIT: For those interested:
$$r^2=x^2+y^2+z^2$$
$$-x^2-y^2 = z^2 -r^2$$
and then sub back in

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