# Homework Help: Magnetic field rod

1. Nov 3, 2009

### tomfrank

1. The problem statement, all variables and given/known data
i have a rod of 35 cm and a current of 10 A. the rod is a paramagnetic aluminum rod. how do i find the magnetic field inside the rod?

2. Relevant equations
the fact that the rod is paramagnetic changes anything?

3. The attempt at a solution

2. Nov 3, 2009

### Troels

Yes. You now have a magnetization inside the rod that prohibits use of the biot-savart law there.

Also, I'm not sure what you mean by "35 cm" - are you refering to the length? If so, you should be very careful in applying ampéres law as you do not have a perfect cylindrical symmetry.

3. Nov 3, 2009

### tomfrank

sorry it is the diameter. and the rod is consider to to infinitely long.

so what should I use?

4. Nov 3, 2009

### Troels

Okay - that eases thing up a bit :)

Use Ampéres law for the H-field:

$$\oint_{\partial \mathcal{S}}\vec H \cdot d\vec \ell = \int_{\mathcal{S}}\vec J_\mathrm{free}\cdot d\vec a$$

For points outside the wire, this of course reduces to the familiar form:

$$\oint_{\partial \mathcal{S}}\vec H \cdot d\vec \ell = I_\mathrm{free,encl}$$

but for points inside the wire, you must be careful to only include the part of the current that is enclosed by the loop - assume that the current density is uniform, if not stated otherwise.

THen you should find:

$$H\cdot2\pi r = J_0 \pi r^2 \quad\Rightarrow\quad H=\frac{J_0 r}{2}$$ for r < 17.5 cm

Where J0 is the totalt current divided by the crossectional area of the wire

(take a moment to verify this)

Now use

$$\vec B = \mu_0(\vec M + \vec H)$$

together with

$$\vec M = \chi \vec H$$

to find the required result

Last edited: Nov 3, 2009
5. Nov 3, 2009

### tomfrank

so once i find the H which is the magnetic field i just plugged in the
$$\vec B = \mu_0(\vec M + \vec H)$$
along with
$$\vec M = \chi \vec H$$

so is $$J_0$$
the current they gave me?

6. Nov 3, 2009

### Troels

As I stated, it is the current density which is the total current divided by the cross-sectional area of the wire. THat is:

$$J_0=\frac{10\,\mathrm{A}}{\pi(0.35 \,\mathrm{m})^2}$$

7. Nov 5, 2009

### tomfrank

how will I find the bound current density Jb and Kb.

Jb= curl M

Kb= cross product of M and n.

is this right and how do I do the curl?

8. Nov 5, 2009

### Troels

It's right - so just plug-in the magnetization you got from $$\vec M = \chi \vec H$$