# Homework Help: Coil current physics homework

1. Mar 29, 2006

### pkossak

I was wondering about the following problem:

You are looking down on a single coil in a constant magnetic field B = 0.9 T which points directly into of the screen. The dimensions of the coil go from a = 6 cm and b = 15 cm, to a* = 20 cm and b* = 19 cm in t=0.028 seconds. If the coil has resistance that remains constant at 1.7 ohms, what would be the magnitude of the induced current in amperes?

Now, I have the answer, and I was told how to get it. I used the formula I = (delta A*B)/(delta t*R)

What I was wondering was if someone could tell me what rule or law this formula came from? I can't figure out how to derive it from any of the formulas given in this chapter. Thanks a lot.

2. Mar 29, 2006

### pkossak

sorry I just realized this was the wrong place to post this question!

3. Mar 29, 2006

### da_willem

It can be derived using one of the Maxwell equations: Faraday's law:

$$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$

or in integral form

$$\int \vec{E} \cdot d \vec{l} = -\frac{d}{dt} \int \vec{B} \cdot d \vec{A} = -\frac{d \Phi}{dt}$$

With phi the flux B*A enclosed. The first integral is equivalent to (minus) the voltage across the loop so:

$$V = \frac{d \Phi}{dt} = A \frac{d B}{dt} +B\frac{d A}{dt}$$

Or in your case only the last term is nonzero, so

$$I=\frac{V}{R}=\frac{dA}{dt} \frac{B}{R}$$

4. Mar 29, 2006

### Hootenanny

Staff Emeritus
I'm afraid I don't agree with you here. You are effetively saying that the induced emf is equal to the induced current . Also Faraday's law is rate of change of flux, therefore the BA should be enclosed.

-Hoot

5. Mar 29, 2006

### da_willem

As far as I made a mistake, you're not too good at explaining where.

Where? I said I=V/R which is just Ohms law

Ofcourse, I also stated that explicitly

So I'm sorry but I can't really see what's wrong with my derivation...?!

6. Mar 29, 2006

### Hootenanny

Staff Emeritus
Sorry, It just confused me when the flux was in a different fraction. The last point is minor but you say;

$$I=\frac{V}{\not R}=\frac{dA}{dt} \frac{B}{\not R}$$
$$I = V = \frac{d(AB)}{dt}$$

7. Mar 30, 2006

### da_willem

Thanks for your reply. Do you mean to say I=V/R should be I=V?

This is not only in conflict wih Ohm's law, it also yields the wrong result. It is the emf that is equal to (minus) the time derivative of the flux, not the current...

8. Mar 30, 2006

### Hootenanny

Staff Emeritus
No, look at the equation you have written;

The two resistances would cancel, leaving you with;

$$I=V=\frac{dAB}{dt}$$

Which implies that I = V. I knew what you meant, but at first glance it may be confusing.

9. Mar 30, 2006

### da_willem

I'm totally lost in what you mean...

You can't just cancel the R's in the last equality and do noting with the first equality...

$$I=\frac{V}{R}=\frac{dA}{dt} \frac{B}{R}$$

Multiplying by R

$$IR=V=\frac{dA}{dt} B$$

So IR=V (Not I=V) which is just Ohm's law again!

10. Mar 30, 2006

### Hootenanny

Staff Emeritus
Ahh sorry, my mistake I was veiwing them as two seperate equations, my apologies da_willem.