Why Doesn't the Coil Radius Affect the Induced EMF Calculation?

[Ithryndil]

1. Homework Statement

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. Homework 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
$$Therefore:<br /> Emf = NA\mun(I)$$ ...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.

 

[Ithryndil]

Bump.

 

[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.

 

[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:

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

 

[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)

 

[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.