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

  • #1
Hello again everyone

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

B(r) = [tex]\frac{1}{2}\mu_0[/tex] (J x r)

from Ampere's law:

[tex]\nabla[/tex] x B = [tex]\mu_0[/tex] 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
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,024
578
It's not checking

[tex] \nabla \times \left(\frac{1}{2}\mu_{0} \vec{J}\times\vec{r}\right)\neq \mu_{0}\vec{J} [/tex]
 
  • #3
Tell me about it. I'll copy the full question out, see if anything flies.

Use of Ampere's Law

A long circular rod of radius R, made of conducting material, has a cylindrical hole of radius a bored parallel to its axis and displaced from the centre of the rod by a distance d. The rod carries a current I distributed uniformly over its cross-section.

Consider the superposition of two currents flowing in opposite directions, and hence show that the magnetic field in the hole is uniform and equation to:

B = [tex]\frac{\mu_0 d I}{2 \pi (R^2 - a^2)}[/tex]

[Hint: Deduce from Ampere's law that B(r) = [tex]\frac{1}{2}\mu_0[/tex](j x r) for each current, where j is the current density, and determine the appropriate origin of r].

First part, finding J, is easy enough. I = j.ds = j[tex]\pi (R^2 - a^2)[/tex] gives j in terms of I as required for the final equation.
 
Last edited:
  • #4
Jesus, the number of times I've had to edit that post.
 
  • #5
It's not checking

[tex] \nabla \times \left(\frac{1}{2}\mu_{0} \vec{J}\times\vec{r}\right)\neq \mu_{0}\vec{J} [/tex]
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) = [tex]\frac{1}{2}\mu_0[/tex](J x r)

then:

[tex]\nabla[/tex] x B = [tex]\frac{1}{2}\mu_{0} \nabla[/tex] x (J x r)

= [tex]\frac{1}{2}\mu_0[/tex]( J ([tex]\nabla[/tex].r) - r ([tex]\nabla[/tex].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 [tex]\nabla[/tex].j = 0:

[tex]\nabla[/tex] x B = [tex]\frac{1}{2}\mu_0[/tex]( J ([tex] \frac{1}{r}\frac{d}{dr}r^2 [/tex]))

= [tex]\frac{1}{2}\mu_0[/tex] * 2J

= [tex]\mu_0[/tex]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:

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

Replies
3
Views
2K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
16
Views
7K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
18
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
5K
Top