1. Oct 30, 2008

Ithryndil

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

A coil of 15 turns and radius 10cm surrounds a long solenoid of radius 2cm and 1x10^3 turns/m. The current in the solenoid changes as I = (5A)sin(120t). Find the induced emf in the 15-turn coil as a function of time.

2. Relevant equations

emf = -Nd[flux]/dt

$$\Phi = \oint[B*dA]$$

3. The attempt at a solution

Edit: I figured out what I was doing wrong. However, I am still puzzled on one aspect of this problem

When I figured out what I did wrong I came up with:

$$B = \munI$$
$$Emf = \frac{d}{dt}(NBAcos\theta) = \frac{d}{dt}(NBA)$$ cos(theta) = 1 in this case
$$Therefore: Emf = NA\mun(I)$$ ...and we have I so it's a simple derivative from there

What I don't get is why the .1m radius doesn't really come into play. Unless I missed something or am not thinking here at all, it's not necessary for this problem, other than to tell us the coils are outside the solenoid.

Last edited: Oct 30, 2008
2. Oct 30, 2008

Ithryndil

Bump.

3. Oct 30, 2008

nasu

What area is the one you call A? The area of the solenoid i suppose.
When you calculate the flux through the coil you should use (in general) the area of the coil.
But... for a "long solenoid", the field on the outside of the solenoid is very weak and is usually neglected in elementary problems.
The field of the solenoid B=I*N/L (no area) is inside the solenoid - does not depend upon the area of the solenoid.

4. Oct 30, 2008

Ithryndil

The A I am using is the area of the solenoid...at least the cross sectional area.

Also, let me correct a mistake in that last equation:

$$Therefore: Emf = NA(\mu)n\frac{d}{dt}(I)$$

5. Oct 30, 2008

nasu

Then you should see why.
When you calculate the flux through something, you should use the area of the "something".
Here you use the area (cross-section) of the solenoid instead because the field outside the solenoid is neglected (and you don't have a formula for the field outside anyway, in elementary textbooks)

6. Oct 30, 2008

Ithryndil

Ok, so it doesn't matter that the coil of 15 turns is .08 cm away from the solenoid, we only need the solenoid radius for the flux...hmm, I must have missed that in class.