Finding Magnetic Field in Paramagnetic Aluminum Rod of 35 cm, 10A

[tomfrank]

Homework Statement

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?

Homework Equations

the fact that the rod is paramagnetic changes anything?

The Attempt at a Solution

 

[Troels]

the fact that the rod is paramagnetic changes anything?

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 referring to the length? If so, you should be very careful in applying ampéres law as you do not have a perfect cylindrical symmetry.

Please clarify

 

[tomfrank]

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

so what should I use?

 

[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

 

[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?

 

[Troels]

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

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

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

 

[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?

 

[Troels]

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?

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