# Homework Help: Magnetic Induction (Non-uniform Current)

1. Aug 24, 2013

### roam

1. The problem statement, all variables and given/known data

long wire of radius R0 carries a current density j given by

[Broken]​

Find the magnetic induction B inside and outside the wire.

2. Relevant equations

Current density: $J=\frac{I}{A}$

Ampere's law: $\oint B.dl = \mu_0 I_{enc}$

3. The attempt at a solution

For the magnetic field inside (ρ<R0) using Ampere's law:

$\oint B . dl = B 2 \pi \rho = \mu_0 I_{enc}$

Now I'm not sure what to use as Ienc. If I use the relationship

$I_{enc}=JA=j_0 \frac{\rho}{R} \pi \rho^2 = j_0 \frac{\rho^3 \pi}{R}$

I get

$B= \frac{\mu_0 j_0 \rho^2 }{2 R_0}$

But if I integrate (in cylindrical coordinates) I will get a different value for Ienc:

$I_{enc}= \int \frac{j_0 \rho}{R_0} (\rho \ d \rho \ d \phi) = \frac{j_0}{R_0} 2 \pi \int^\rho_0 \rho^2 d \rho = \frac{2 \pi j_0 \rho^3}{3R_0}$

Therefore I get a different value for B:

$B=\frac{\mu_0 j_0 \rho^2}{3R_0}$

So which method is correct?

And for the magnetic field outside, I get $B=0$ since the RHS of Ampere's equation is 0. But shouldn't the magnetic field inside a conducting wire be zero, and non-zero outside it?

Any explanation would be greatly appreciated.

Last edited by a moderator: May 6, 2017
2. Aug 24, 2013

### davidchen9568

Because J is a function of radius. $$I=\int { J\cdot dA }$$ The second method looks fine.

For the second question: What is the enclosed current outside? Is it zero?

3. Aug 28, 2013

### rude man

You don't need to integrate over phi explicitly. Using r in place of ρ:

A differential annulus of area dA = 2πr dr contains current di = j(r)dA = 2πr j(r) dr
so current within radius r = 2πj0/R00r r2 dr
= 2πj0r3/3R0
= 2πrB/μ0 etc.