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Faraday's law of mutual induction

by darksyesider
Tags: faraday, induction, mutual
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May6-14, 06:41 PM
P: 55
When solving for mutual induction, how do you know what surface to take the flux over?
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May6-14, 08:11 PM
P: 160
You are allowed to choose any surface* as long as its boundary is the loop around which you are computing the emf. In general you choose a flat surface if that's possible, so as to make calculations easier, but you could choose whatever surface you like as long as it has the right boundary.

* okay, not quite any surface -- you have to choose an orientable surface, but you don't need to worry about that unless you are deliberately being perverse.
May6-14, 08:19 PM
P: 55
So you ALWAYS make the boundary the loop where you're computing the emf?
Can you think, by any chance, of an example of this? (besides coaxial cables which i made a post about a few days/hours ago)

May6-14, 08:46 PM
P: 160
Faraday's law of mutual induction

Yes, the loop around which you are computing the emf must always be the boundary of the surface.
If you are confused about this, I would suggest that you watch Walter Lewin talk about it (start at 5:00) if you haven't already.

As for an example, basically any Faraday's Law problem will do the trick.

Here's an example: A circuit consisting of a circular loop of wire (radius 1 cm) and an LED is placed inside a solenoid, with the loop of wire concentric with the coils of the solenoid. The solenoid is turned on and the magnetic field inside smoothly increases to 0.1 T over 0.001 seconds. The LED has a 1 ohm resistance and will explode if a current greater than 50 mA flows through it. Does the LED survive?

Answer: We want to calculate the EMF of the circuit, so the circuit has to be the boundary of my surface. I choose the flat surface, which is just a circle, and in this case I am in luck -- the magnetic field is perpendicular to that surface. So the magnetic flux is BA. The rate of change of magnetic flux is 0.1 T * pi(0.01 m)2/0.001 s = 0.01*pi Tm2/s. Therefore, the emf is 0.031 V, which produces a current of 31 mA. That is not enough to explode the LED.

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