# Average emf is it same as induced emf?

1. Jul 19, 2006

### skull42484

average emf is it same as induced emf??

Q: a 25-turn circular coil of wire has diameter 1.00m. it is placed with its axis along the direction of the earth'smagneticfield of 50.o uT, and then in 0.200s it is flipped 180 degrees. an average emf of what magnitude is generated in the coil?

N = 25
r = .5m --------> A = 0.785m square
B = 50.0uT
t = .200s

emf = [25 * 0.785 *50 * 10^-6 * (cos180)] / .200

am i on the right track

2. Jul 19, 2006

### Office_Shredder

Staff Emeritus
The reason they ask for the average EMF, is because it actually does change depending on at what angle in the 180 degree flip it is at (when it's halfway vs. 2/3s of the way flipping, for example). But since that's tough to calculate, they use average, so all you need to do is plug and chug using the formula

3. Jul 19, 2006

### Andrew Mason

Can you explain what law you are using and write out the formula you are applying? I think that it is asking for the emf that results from the average rate of change of flux (total change in B/total time).

AM

4. Jul 19, 2006

### skull42484

i am using the faraday's law of induction

induced emf = -N (d/dt) (magnetic flux)

the magnetic flux through the loop of area is
magnetic flux = BA cos__ (the degree between the magnetic feild and the loop of area)

then induced emf = -N (d/dt) (BAcos__)

5. Jul 19, 2006

### Office_Shredder

Staff Emeritus
The only problem is your change in magnetic field... it's going from 50 to -50 microTeslas, not from 0 to -50

6. Jul 19, 2006

### skull42484

so the angle is not involved to the equation

and the equation that i should use is
emf = NA*(the change in magnetic field) / (the change in time)
= (25 * 0.785 * 100*10^-6) / 2
am i on the right track this time

7. Jul 19, 2006

### Office_Shredder

Staff Emeritus
Except it's .2 seconds, not 2 seconds

Other than that, it looks good

EDIT: Not to confuse you, the angle is critical in the question. While you don't directly use it in the calculation, you do need to know it to calculate the beginning and ending flux.... for example, try now doing the same problem, where the loop is only rotated 90 degrees in .2 seconds

Last edited: Jul 19, 2006
8. Jul 19, 2006

### skull42484

if is 90 degrees, then the change in magnetic field is 50
if so the emf = (25 * 0.785 * 50*10^-6) / .2

9. Jul 19, 2006

### Office_Shredder

Staff Emeritus
Exactly. So you can see how the angle is important, it's just not actually put into the final equation.

Awesome