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Induced emf in a Rectangular Coil

  • Thread starter purduegirl
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  • #1
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Homework Statement



A rectangular coil of 150 turns has dimensions of 23 x 70 cm and is located in a uniform 3 T magnetic field. In 1.2 s, the plane of the coil is rotated from a position where it makes an angle of 15° with the magnetic field to a position where it makes an angle of 72°. Calculate the average emf induced in the coil.


Homework Equations



E = -N( Change in Flux/ change in time)
Flux = BAcos theta


The Attempt at a Solution



To find the change in flux, I set up the equation of Change of flux = BAcos72 -BAcos15. For this I got -.3172 T*m^2.

I plugged this into the first equation = -150 (-.3172/1.2) I got 39.66 N*m/C.
 

Answers and Replies

  • #2
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I'm not sure about the math but your problem-solving steps look fine...
 
  • #3
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I tried this once again, here's my math.

Flux = NBAcos[tex]\theta[/tex]
Flux = (3.0 T)(.23 m*.70 m)cos(72-15)
Flux = 3T( 0.161 m^2)cos(57)
Flux = 0.2630606539

Emf = -N( Flux/ time change)
Emf = -150( 0.2630606539 V*s / 1.2 s)
Emf = - 150( 0.2192172116 V)
Emf = -32.88 V

This and the postive answer is wrong.
 
  • #4
alphysicist
Homework Helper
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Hi purduegirl,

Homework Statement



A rectangular coil of 150 turns has dimensions of 23 x 70 cm and is located in a uniform 3 T magnetic field. In 1.2 s, the plane of the coil is rotated from a position where it makes an angle of 15° with the magnetic field to a position where it makes an angle of 72°. Calculate the average emf induced in the coil.


Homework Equations



E = -N( Change in Flux/ change in time)
Flux = BAcos theta


The Attempt at a Solution



To find the change in flux, I set up the equation of Change of flux = BAcos72 -BAcos15. For this I got -.3172 T*m^2.
I don't think these angles are correct. In the flux equation, the angle theta is the angle between the magnetic field and the normal direction to the loop. If the field is making an angle of 15 degrees with the plane of the loop, what angle is it making with the normal to the loop?
 
  • #5
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Would it be 90-15 since the magnetic field is perpendicular to the velocity?
 
  • #6
alphysicist
Homework Helper
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Would it be 90-15 since the magnetic field is perpendicular to the velocity?
That numer is right, but the reasoning is wrong. The angle in the formula is the angle that the magnetic field makes with the normal to the loop.

If the magnetic field is passing straight through the loop, the angle is 0 degrees or 180 degrees. If the magnetic field is "hitting" the loop edge on, the angle would be 90 degrees.

This problem says that the B-field is 15 degrees away from hitting the loop edge on, so it's (90-15) degrees away from passing straight through the loop.
 
  • #7
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But, would that also apply to the 72 degrees? I talked to my T.A. about this problem yesterday and he told met that I should divide the cos theta to get the answer. Would this be the correct reasoning?
 
  • #8
alphysicist
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But, would that also apply to the 72 degrees? I talked to my T.A. about this problem yesterday and he told met that I should divide the cos theta to get the answer. Would this be the correct reasoning?
If I'm visualizing it correctly, then it should also apply to the 72 degrees.

What do you mean by divide the cos theta?
 
  • #9
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My T.A said that I should divide the cos of angle 1 by the cos of angle 2. If I apply the subtract from 90 rule to both angles, I still get the difference is 57 degrees as I did before with the regular angles.
 
  • #10
alphysicist
Homework Helper
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My T.A said that I should divide the cos of angle 1 by the cos of angle 2. If I apply the subtract from 90 rule to both angles, I still get the difference is 57 degrees as I did before with the regular angles.
I'm sorry, I still don't understand what was meant by dividing the cosines. I mean, the cosine of one of the angle could have been zero, so I don't see how dividing the cosine could be okay. Do you have an example?



The difference in the angles is not important; it's the difference in the cosines of the angles, as you have in your first post. In your first post, you had:

Change of flux = BAcos72 -BAcos15
This is okay, except the angles should be 18 and 75.
 
  • #11
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I tried this once again, here's my math.

Flux = N(BAcos 18 - BAcos 75)
Flux = N [(3.0 T)(.23 m*.70 m)cos(18)-(3.0 T)(.23 m*.70 m)cos(75)]
Flux = -150 (.482[(.95106) - (.25882)])
Flux = -150 ( .3343506986)/1.2 s
Average Induced Current = 41.79 V

Everytime, I do this problem my number changes, what am I doing wrong!
 
Last edited:
  • #12
alphysicist
Homework Helper
2,238
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I tried this once again, here's my math.

Flux = N(BAcos 18 - BAcos 75)
Flux = N [(3.0 T)(.23 m*.70 m)cos(18)-(3.0 T)(.23 m*.70 m)cos(75)]
Flux = -150 (.482[(.95106) - (.25882)])
Flux = -150 ( .3343506986)/1.2 s
Flux = 41.79 V

Everytime, I do this problem my number changes, what am I doing wrong!
I think that last number should be the average induced emf you're looking for. Are you saying that's the wrong answer?
 
  • #14
dynamicsolo
Homework Helper
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All right, I was about to type one thing, and am going to retract it. I see nothing wrong with your calculation.

Are you doing this for a computer-based problem system, like WebAssign? I can think of two possibilities, both of which I've seen in practice.

The scary one is that the person setting up the problems didn't set a tolerance percentage for the answer. The default is zero, which means you must match the computer's calculations to eight or sixteen decimal places (good luck with that!).

The less scary one is that the person who entered the answer did the problem wrong. (I've seen that happen more than once.) One thing you might try is finding the answer with the "wrong" angles, 15º and 72º; the person working this out and entering the answer to be matched could have misunderstood the description of the situation.

Oh, one other thing to try: include the minus sign in your input.

If that doesn't work, see if anyone else is having the same trouble. If no one seems to be getting the computer to accept an answer, there could well be something wrong with the answer the system was given. You'd then take it up with the course instructor. (When this has happened -- not often -- in the past, it has turned out to be an error on their end...)
 
Last edited:
  • #15
alphysicist
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Yes, it is.
Based on the numbers in your problem, I believe that is the magnitude of the average induced emf, unless I'm just not reading it correctly somehow.

I noticed you dropped the minus sign from your formula. Do you think instead of the magnitude, they want you to keep the sign?
 
  • #16
alphysicist
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Hi dynamicsolo,

OK, I see one problem. This line should be right:

Flux = N [(3.0 T)(.23 m*.70 m)cos(18)-(3.0 T)(.23 m*.70 m)cos(75)]

But in going to this next line,

Flux = -150 (.482[(.95106) - (.25882)]) ,

you put in the value for N, which is 150, and dropped the value for B. There should be another factor of 3 here:

Flux = -150 · 3 · (.482[(.95106) - (.25882)])

I don't think that's it; the .482 includes the area and the magnetic field, I believe.
 
  • #17
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Yes, it does include the magnetic field of 3.
 
  • #18
alphysicist
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purduegirl,

Unless they want you to keep the minus sign I mentioned (so the answer is -41.79 V), I don't see what else could be wrong. Either I'm reading it wrong somehow, or perhaps the numbers were copied wrong, or maybe it's right and whatever is checking the answer is incorrect.
 
  • #19
dynamicsolo
Homework Helper
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Yes, it does include the magnetic field of 3.
Yeah, sorry. I was having trouble reading a couple of the lines. I've since altered my post. Try including the minus sign on the answer, and have a look at my post now...
 

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