# Induced emf in secondary coil - help

1. Oct 20, 2008

### jcvince17

induced emf in secondary coil - help ASAP Please

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

A very long, straight solenoid with a cross-sectional area of 6.00 cm2 is wound with 40.0 turns of wire per centimeter, and the windings carry a current of 0.250 A. A secondary winding of 2.00 turns encircles the solenoid at its center. When the primary circuit is opened, the magnetic field of the solenoid becomes zero in 5.00×10−2 s .

What is the average induced emf in the secondary coil?

2. Relevant equations

E = Delta flux / delta time

flux = BA cos Theta

B = U0*n*I

3. The attempt at a solution

I have no idea where to begin. to find the flux I need the Magnetic Field. I do not know the magnetic field value so How can I find flux to in turn find E?

I think I have found B on the first coil.
B = U0*n*I
B = U0*240*0.250A I used 240 as it is 40 turns per cm.

B = 7.54*10-5 ???

then if i fund flux of coil 1 =

flux = BA
fulx = 7.54*10-5 * .0006 m2 = 4.53*10-8 ???

i tried doing the same for the other coil, but I do not have anything to get area of that coil.

i used:

N1/N2 = I1/I2 for I2 and got I2 = .0125A ????

Last edited: Oct 21, 2008
2. Oct 21, 2008

### alphysicist

Re: induced emf in secondary coil - help ASAP Please

Hi jcvince17,

I don't see where you got 240 from here. They say that n=40 turns per cm, and you need n to have units of turns per meter.

3. Oct 21, 2008

### jcvince17

when i convert 6 sq cm to sq meters i get .0006. so do i take the square root of that and use .025 m x 40 turns and get .98 = n?

i am lost on this. it is an online hw assignment and we havent even began this chapter in class yet.

4. Oct 21, 2008

### alphysicist

No, n is just the number of turns that the solenoid has per meter. They give you n directly--it is 40 turns/cm. However, you need to do a units conversion to turn that into turns/m.

So think of a coil. If I say there are 40 turns (loops) in a 1cm long coil, how many turns are there in a 1m coil (assuming everything is identical except the length)?

5. Oct 21, 2008

### jcvince17

40/1cm = .4/1m

correct?

6. Oct 21, 2008

### alphysicist

No, you went the wrong way with the decimal. If there is room for 40 turns in a small 1cm length of coil, a very long 1m coil would have room for many more turns (not less).

7. Oct 21, 2008

### jcvince17

oops you are right. sometimes I do not think with common sense or to the obvious.

40/1cm = 4000/1m

8. Oct 21, 2008

### jcvince17

ok so if n = 4000 turns.

B of coil 1 = B = U0*n*I

B = U0*4000*0.250 = 1.25*10-3

therefore flux of coil 1 = BA cos theta

flux 1 = 1.25*10-3 * .0006 m 2 = 7.54*10-7

correct thus far? i assume B is perpendicular since no angle is mentioned.

am i correct so far?

what and how do i do coil 2?

9. Oct 21, 2008

### alphysicist

That looks right to me.

This is fine for the flux through one loop of coil 1; however, you don't need this. You just need to be able to find the flux through coil 2. You do it the same way as coil 1.

Once you have that, you can determine the induced emf.

(By the way, your two equations in the original post:

E = Delta flux / delta time

flux = BA cos Theta

would give you the induced emf magnitude for the case when the secondary loop only has one turn. When the secondary loop has N turns, you'll need a factor of N on the right hand side of one of these equations.)

10. Oct 21, 2008

### jcvince17

I fixed the equation.

now I have done this:

N1/N2 = I1/I2

4000/200 = .250/I2

solve for I2 = 1.25*10-2 ??

B for coil 2 = Uo*n*I
B = Uo*200*1.25*10-2 = 3.15*10-6 ???

how do i find the area of the second coil? or is it assumed the same as the first coil but with fewer loops?

if i assume same area of .0006 sq meters i get:

flux = BA

flux = 3.15*10-6 * .0006 = 1.89*10-9 (multiply by 2 or 200 for number of turns?)

then take this answer and divide by delta t = emf?

i know what the answer is supposed to be but have no idea how to get there. i edited my original post to state the actual question. i need the avg emf for the secondary coil. just for clarification.

Last edited: Oct 21, 2008
11. Oct 21, 2008

### alphysicist

This is not correct (and not needed for this problem).

When we talk about the flux through coil 2, we are saying that coil 1 produces a B field, and we want to know the flux from that field that goes through coil 2.

flux = B A

here flux is the flux through coil 2 that we are looking for, B is the field from coil 1 (not coil 2!), and A is the area of the field captured by coil 2.

So in this case the area where the B field is nonzero is equal to the area of coil 1, so that is what goes here.

Right, once you find the change in flux you will divide by the time interval for that change. (Also remember you have to use the number of loops N of coil 2 somewhere in the induced emf calculation.) That should give you the induced emf. Do you get the answer?

12. Oct 21, 2008

### jcvince17

ok so flux of 2 = BA

flux = 1.25*10-7 (from B1)*.0006 = 7.5*10-11 right?

now which flux do i need to get my "change" in flux? once i find this i can divide by .05 (delta t) and then multiply by 2 (number of turns) and hope to have my answer. the supposed answer is 30.2*10-6 V

i am very sorry for all my mistakes and questions. but i am trying to teach this to myself with your help.

13. Oct 21, 2008

### alphysicist

Maybe I'm getting lost in the details here, but should B here be 1.25*10-3?

When you find this flux, this will be the flux through coil2 at the initial time (when the current in coil1 is 0.25A).

Next repeat the calculation for the final time: they say after a certain period of time the B field from coil1 is zero, so at that point what is the flux through coil2?

The difference in the final and initial flux is the change in flux.

Once you find the right change in flux, that should give the right answer.

14. Oct 21, 2008

### jcvince17

yes.

flux initial = 1.25*10-3 * .0006 = 7.5*10-7

flux final = 0

delta flux = final - initial = 7.5*10-7 - 0 = 7.5*10-7

delta time = 5.0*10-2

emf = (delta flux / delta time) * N

emf = 3*10-5 = 30 micro volts

i got it!!! i had to have my answer in micro volts so i had to enter 30 into the box and i finally got it.

Thank you for all your help. I do appreciate you taking the time to help me work through this. Thanks a million for your guidance and hints.

15. Oct 21, 2008

### alphysicist

That looks great! And you're welcome, I'm glad to help.