# Deriving B(r) from Ampere's law - For tomorrow (02/23)

1. Feb 22, 2007

### El Hombre Invisible

Hello again everyone

Part of a problem I've been set is to show that the equation:

B(r) = $$\frac{1}{2}\mu_0$$ (J x r)

from Ampere's law:

$$\nabla$$ x B = $$\mu_0$$ J.

The problem presents no... uh... problem thereafter, but I'm at a loss where to begin. I've been playing around with random vector identities for a while but I'm rubbish at remembering these things, my notes are in mayhem and my textbooks aren't helping. If I could get a hint where to start, that would be great.

Oh, and this is for tomorrow, so quite urgent.

Cheers,

El Hombre

2. Feb 22, 2007

### dextercioby

It's not checking

$$\nabla \times \left(\frac{1}{2}\mu_{0} \vec{J}\times\vec{r}\right)\neq \mu_{0}\vec{J}$$

3. Feb 22, 2007

### El Hombre Invisible

Tell me about it. I'll copy the full question out, see if anything flies.

First part, finding J, is easy enough. I = j.ds = j$$\pi (R^2 - a^2)$$ gives j in terms of I as required for the final equation.

Last edited: Feb 22, 2007
4. Feb 22, 2007

### El Hombre Invisible

Jesus, the number of times I've had to edit that post.

5. Feb 22, 2007

### El Hombre Invisible

Actually, expanding this it might be okay. Could you (or someone) double-check this for me please?

Starting the wrong way round, if:

B(r) = $$\frac{1}{2}\mu_0$$(J x r)

then:

$$\nabla$$ x B = $$\frac{1}{2}\mu_{0} \nabla$$ x (J x r)

= $$\frac{1}{2}\mu_0$$( J ($$\nabla$$.r) - r ($$\nabla$$.j)).

Now since each current is independant of position in the rod and is uniform everywhere, the amount of charge in any given volume element must stay the same, so $$\nabla$$.j = 0:

$$\nabla$$ x B = $$\frac{1}{2}\mu_0$$( J ($$\frac{1}{r}\frac{d}{dr}r^2$$))

= $$\frac{1}{2}\mu_0$$ * 2J

= $$\mu_0$$J,

which is Ampere's law. Not that in a million years I would ever think to do that the other way round, but does it look okay?

Many thanks,

El Hombre

P.S. It'll take me another 20 minutes to sort out the LaTex problems so if this is incomprehensible, make a cuppa and please come back!
P.P.S. Wow! Other than consistently spelling nabla 'nable', I did alright. 'Slike riding a bike.

Last edited: Feb 22, 2007