# Magnetic Flux through 1 loop due to current on the other

1. Aug 8, 2014

### lion_

The scenario is the following, I am given 2 loops with the same radius, r, a distance of d, and same current of I. In the left loop the current goes counter clockwise, in the right loop the current is clockwise. The two loops centers lie on the same axis which are perpendicular to the plane of the loops. I am asked to find the magnetic flux of the left loop due to the current on the right loop.

I know that the magnetic flux of a loop is $$\phi=B\pi r^2$$ where $$B=\dfrac{\mu_0 I}{2R}$$ So how exactly do I find the Total magnetic flux on the loop due to the magnetic flux on the other? Since the current is opposite I will be subtracting the 2 fluxes.

So $$\phi_{self}=\phi_L-\phi_R$$ which is $$\dfrac{\mu_0I}{2r} \pi d^2 - \dfrac{\mu_0I}{2r}\pi d^2=0$$ I dont think this makes much sense to me....

2. Aug 8, 2014

### LucidLunatic

First of all, it does not make sense to ask for the magnetic flux ON something. You can have magnetic flux THROUGH something, but not ON it. Make sure you're not supposed to be calculating the magnetic force on the loop.

Second, since you only care about the flux through the second loop due to the current in the first, you do not need to worry about the magnetic field generated by the second loop. It is out of the scope of the question. However, you will need to know the magnetic field at all points within the second loop caused by the first. As such, you need to:

1. Find the function for the magnetic field caused by the first loop at all points in space.
2. Integrate this over the area of the second loop.

This will give you the total flux through the second loop, caused by the current in the first loop.

3. Aug 8, 2014

### lion_

$$B=\dfrac{\mu_0}{2\pi} \cdot \dfrac{\mu}{l^3}$$

Is this the function you are talking about? Just substiute the numbers and that is it?

4. Aug 9, 2014

### Zondrina

I agree that this seems like a magnetic force type of question.

The magnetic dipole $\vec{\mu}$ of the right loop indicates a net flux $\Phi_B$ through the left loop. You know the flux through the loop is given by:

$\Phi_B = \int \vec B \cdot d \vec A$

Where it's safe to assume the field is uniform if the distance $d$ is small. Otherwise that function you posted should help.

Last edited: Aug 9, 2014
5. Aug 11, 2014

### rude man

No more posts for nearly 2 days? Then I wll hazard the opinion that this is an extremely difficult problem, requiring special functions like Bessel functions and elliptical integrals.