Induced emf in secondary coil - help

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Homework Help Overview

The discussion revolves around calculating the average induced electromotive force (emf) in a secondary coil encircling a solenoid. The solenoid has specific parameters, including a defined cross-sectional area, number of turns per unit length, and current. The problem involves understanding the relationship between the magnetic field generated by the solenoid and the induced emf in the secondary coil when the primary circuit is opened.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the magnetic field (B) using the formula B = U0*n*I, questioning the conversion of turns per centimeter to turns per meter. There are attempts to calculate the magnetic flux through the coils and discussions about the area of the secondary coil. Some participants express confusion about the relationship between the number of turns and the induced emf.

Discussion Status

Participants are actively engaging with the problem, sharing calculations and questioning assumptions. Some have provided guidance on unit conversions and the application of formulas, while others are still seeking clarity on specific steps and concepts. There is a recognition of the need to find the change in flux to calculate the induced emf, but no consensus has been reached on the final approach.

Contextual Notes

Participants note that the problem is part of an online homework assignment and that the relevant chapter has not yet been covered in class, which may contribute to the confusion. There is also mention of the need to consider the number of turns in the secondary coil when calculating the induced emf.

jcvince17
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induced emf in secondary coil - help ASAP Please

Homework Statement



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?

Homework Equations



E = Delta flux / delta time

flux = BA cos Theta

B = U0*n*I

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?

Please get me started correctly.

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:
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Hi jcvince17,

jcvince17 said:

Homework Statement



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 .

Homework Equations



E = Delta flux / delta time

flux = BA cos Theta

B = U0*n*I

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?

Please get me started correctly.

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.

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.
 
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 homework assignment and we haven't even began this chapter in class yet.
 
jcvince17 said:
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 homework assignment and we haven't even began this chapter in class yet.

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)?
 
40/1cm = .4/1m

correct?
 
jcvince17 said:
40/1cm = .4/1m

correct?

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

40/1cm = 4000/1m
 
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?
 
jcvince17 said:
ok so if n = 4000 turns.

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

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

That looks right to me.

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.

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
alphysicist said:
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 (* N)

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.)

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.
 
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  • #11
jcvince17 said:
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 ?

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

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?)

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.

So in your equation

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.

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

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
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 :confused:

i am very sorry for all my mistakes and questions. but i am trying to teach this to myself with your help.
 
  • #13
jcvince17 said:
ok so flux of 2 = BA

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

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).

now which flux do i need to get my "change" in flux?


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 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

Once you find the right change in flux, that should give the right answer.
 
  • #14
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. :approve:
 
  • #15
jcvince17 said:
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. :approve:


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

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