# Electromagnetic Induction

1. Apr 8, 2010

### ronaldoshaky

Hello to all.

What form would the magnetic field (due to the square loop) take? My understanding is that the current is changing, so the magnetic field will be changing.

Last edited: Apr 8, 2010
2. Apr 9, 2010

### jrlaguna

You'd have to solve Ampère's law in order to find the exact values. Anyway, you can have some intuition from the right hand rule. If the current is changing, you have to apply Faraday's law. Hm... I need more data to understand what you want.

3. Apr 9, 2010

### ronaldoshaky

The current in the square loop is $$I = I_0 sin (\omega t)$$. The length of the side of the square loop is $$b$$ I don't see how Ampere's Law would help. Isnt Ampere's Law for magnetostatics, and wouldnt it just be giving the magnetic field around one side of the wire?

4. Apr 9, 2010

### jrlaguna

What do you need exactly? The magnetic field in the whole of space, B(x,t)? (That's too hard.) The magnetic flux? The torque on the loop?

Ampère's law will give you the magnetic field for any stationary current. Yours is not stationary, so it will not do the trick by itself.

5. Apr 9, 2010

### ronaldoshaky

The loop is inside a solenoid, and the plane of the loop is perpendicular to the axis of the solenoid. I am trying to work out the flux through the solenoid, due to the current in the loop.

I know the flux is $$\Phi = N A B$$. I was trying to work out $$B$$. I am ultimately trying to work out the mutual inductance, and then the $$V_{emf}$$ and hence the induced current in the solenoid.

6. Apr 10, 2010

### bjacoby

Since you are only interested in mutual inductance between the two conductors (loops) you don't need to bother with finding the flux at all (which will be difficult). To find mutual inductance all you need is the Neumann equation. It is a double integral over the two conductors, source, and the one where emf is induced. In essence at each point on the solenoid you find the induction from every current element on the source loop taking into account the separation R and the angles between the conductors. You add the contributions from all source elements for each point on the secondary. Hence the double integral. It's all very straightforward and simple, but not especially easy. I presume the solution of a square loop and a helix is possible, but it sure sounds like a job for numeric integration on a computer to me!

7. Apr 10, 2010

### ronaldoshaky

I looked up that equation. I think the amount of work involved, as you mention, could be quite labor intensive.

What about looking at this problem from another perspective. One of the other posters there mentioned Faraday's Law. Could I work out the electric field using the current that is mentioned?