# Homework Help: Simple faraday's law problem (phi = what?)

1. May 8, 2014

### darksyesider

1. The problem statement, all variables and given/known data

A toroid with N windings and radius R has cross sectional radius x (x<<R).
The current running through the wires is given by $I = I_0\sin (\omega t)$.
There is a magnetic field in the center of the toroid.

A loop of wire of radius three times that of the toroid is placed around the cross section of the toroid
(like this, but it's circular, not a rectangle).

Find the induced emf in the circular loop of wire.

2. Relevant equations

E = -dphi/dt
3. The attempt at a solution

$\int B\cdot ds = \mu_0 I\implies B(2\pi R) = N\mu_0 I \implies B = \dfrac{\mu_0 N (I_0\sin\omega t)}{2\pi R}$ (*)

then $\mathcal{E} = \dfrac{- d\phi}{dt} = - \dfrac{d(BA)}{dt} = -\pi x^2 \dfrac{dB}{dt}$, where dB/dt is just the derivative of (*).

My question is, that should it be $-\pi x^2$ (the cross section of the toroid) or should it be the area of the loop ( pi*(3x)^2) ?
My reasoning for pi*x^2 is that since there is no flux at parts of the circular loop, it's just the area of the cross section (no flux since no magnetic field!).

2. May 8, 2014

### rude man

You need to integrate B across the toroid's cross-section to the the total flux. B is a function of the radial distance from the toroid's center (not the cross-section center; I mean the center of the hole). You have oversimplified the problem by using the value of B at the outermost radial distance R.

3. May 9, 2014

### darksyesider

EDIT: I don't see why that is true, isn't R fixed?

Last edited: May 9, 2014
4. May 9, 2014

### rude man

Sure is, but the B field isn't confined to r = R.

5. May 9, 2014

### darksyesider

can you specify where it is incorrect? I still don't see what you are referring to.

6. May 9, 2014

### rude man

As I said, you have determined the B field for r=R but have done nothing to determine what it is for r<R. R is the OUTER radius of the toroid. Ir also has an inner radius, which is R-2x. So we have R-2x < r < R.

B is a function of the radius r: B = B(r). You have computed B(R) only. Figure out why B(r) > B(R), r < R, just the way you found B(R).

7. May 10, 2014

### darksyesider

Is there an elementary way to solve this then? (AP Physics C level)?

If the cross section was a square, it would be easy, but here it's harder.

8. May 10, 2014

### rude man

In either case it requires integrating strips of the cross-section, from r = R-2x to r=R, to get the total flux which is what you need.
.
EDIT: ON SECOND THOUGHTS, since you were given x << R your original post solution is close to correct and probably what was intended.

I don't understand how you came up with your alternative area pi(3x)^2. What has a radius 3x???

Last edited: May 10, 2014
9. May 10, 2014

### darksyesider

The larger loop of wire.

10. May 10, 2014

### rude man

Oh, right. But the area you want is, as you said, where the flux is, which is in the toroid cross-section only. Good work and sorry if I sidetracked you. If R >> x had not been given you would have had to integrate.