- #1

TheBigDig

- 65

- 2

- 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
- [tex]\vec{B} = \nabla \times \vec{A}[/tex]

[tex]\vec{A} = \frac{\mu}{4\pi}\frac{m_z}{r^3} (-y,x,0)[/tex]

[tex]\vec{B} = \frac{\mu}{4\pi}\frac{m_z}{r^5} (3xz, 3yz, 3z^2-r^2)[/tex]

[tex]\frac{\partial}{\partial x} \frac{1}{r^3} = -\frac{3x}{r^5}[/tex]

So I was able to do out the curl in the i and j direction and got 3xz/r

[tex]\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}}[/tex]

[tex]\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}}[/tex]

which when added together gives me

[tex]\frac{-x^2-y^2+2z^2}{(x^2+y^2+z^2)^\frac{5}{2}}[/tex].

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

^{5}and 3yz/r^{5}as expected. However, when I do out the last curl, I do not get 3z^{2}-3r^{2}. I get the following[tex]\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}}[/tex]

[tex]\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}}[/tex]

which when added together gives me

[tex]\frac{-x^2-y^2+2z^2}{(x^2+y^2+z^2)^\frac{5}{2}}[/tex].

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