# Strange circuit

1. Apr 23, 2013

### Pi-Bond

1. The problem statement, all variables and given/known data
Electrons in a ferromagnet whose spins are oriented in the direction of, or opposite to, the internal magnetisation carry independent currents I+ and I. This leads to the material behaving as though it has different conductivity σ+ and σ for each of the two current components. These currents may be thought of as flowing through parallel resistances. Two ferromagnetic layers with opposite magnetisation are placed next to each other as shown in the figure.

Each layer has a thickness t and area A. When electrons pass from one layer to the other
their spin-direction remains unchanged. A voltage U is placed across the layers in series
with an external resistor R.

1. Show that the total resistance of the circuit is
$R_0 = \frac{t}{2A} (\frac{1}{\sigma_+} + \frac{1}{\sigma_-})+R$

2. If an external magnetic field above a certain strength is applied to the system, the two ferromagnetic layers will be magnetised in the same direction. Show that the total resistance is now
$R_H = \frac{2t}{A(\sigma_+ + \sigma_-)} +R$

2. Relevant equations
Resistivity
$\rho=\frac{1}{\sigma}$
Resistance
$R=\rho\frac{l}{A}$

3. The attempt at a solution
At first I though that the equivalent circuit for the "component" in the middle would be a parallel one with resistances due to σ+ and σ. In this case the equivalent resistance would be

$R_{eq} = \frac{t}{A}(\frac{1}{\sigma_+ + \sigma_-})$

But this is evidently wrong. Then I thought that each of the two layers has the parallel configuration, and these two layers are in series. In this case the equivalent resistance would be twice the above, which is still wrong.

So it is clear I have the geometry wrong. Can anyone explain the correct geometry?

Last edited by a moderator: Apr 24, 2013
2. Apr 23, 2013

### Staff: Mentor

The metallic part has two indepedent ways where current can flow. One for spin up, and one for spin down. In each part, you have a series of two resistors, corresponding to the left and right side of the metal.
Can you draw this as circuit diagram?

3. Apr 23, 2013

### Pi-Bond

Do you mean something like this?

4. Apr 23, 2013

### Staff: Mentor

Right.

I wonder if that effect was discovered by accident (and explained by QM afterwards), or predicted by QM and measured afterwards.

5. Apr 23, 2013

### Pi-Bond

So are resistors on one side of this parallel geometry having the same conductivity?

By the way this is from a comprehensive exam. Sometime we do see "historical question", but the effects are not really named!

6. Apr 24, 2013

### Staff: Mentor

Depends on the interpretation of "side" and the directions of the magnetic field, probably not.

:(

7. Apr 24, 2013

### Pi-Bond

I figured out the answers. For the first part, both branches had the two resistors of either conductivity in series. For the second part, one branch had the two resistors with σ+ in series, while the other had the two resistor with σ- conductivity in series. I'm not sure why this is the case. Obviously it is due to the magnetisation situation. Can you explain how the magnetisation affects the flowing electrons?

8. Apr 24, 2013

### Staff: Mentor

It is given in the problem statement:
You get one current path with spin-up-electrons (going through both materials) and one current path with spin-down-electrons (going through both materials).

9. Apr 24, 2013

### Pi-Bond

But in the first case, one branch of the parallel circuit has two resistors of σ+ and σ- in series.

In the second case one branch has two σ+ resistors in series, while the other has two σ- resistors in series.

Why this change?

10. Apr 24, 2013

### Staff: Mentor

In the first case, the direction of magnetization is different: electrons seeing σ- in the first material will see σ+ in the second and vice versa.
In the second case, the directions are the same.