Induced current

  • Thread starter SwMarc
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  • #1
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Here is a question I have on a practice exam that I am having a littel trouble with.
A plane loop of wire consisting of a single turn if radius 5cm is perpendicular to a magnetic field that increases uniformly in magnitude from 0.5T to 2.5T in 1.5 seconds what is the resluting indiced current if the coil has a resistance of 2ohms?

I am not sure where to procede I need something that leads me to current so finding a voltage would be nice because I already have the resistance.
perhaps using faradays law?
 

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  • #2
siddharth
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SwMarc said:
perhaps using faradays law?
Yes. Can you post your attempt?
 
  • #3
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Just use
[tex]\epsilon = \frac{d\phi}{dt}[/tex]
 
  • #4
siddharth
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Come on arun, let the original poster show some work!

There's no fun in giving away the answer just like that, is there? The original poster will learn more if he/she figures it out for themselves don't you think :smile:?
 
  • #5
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Well you already said that Faraday's laws are the way to go. I just wrote the mathematical form.
Besides I don't think there is much to do in this problem, otherwise I usually look at the working.

Cheers,
Arun
 
  • #6
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OK I am back. It turns out this problem was on the test with one variation the radius went to 15cm. I used faradays law but got the answer wrong here is what I did.

emf=(pie(.15^2)/1.5=.0471 volts

and then to find the current I=V/R=.0471/2=.0236
 
  • #7
nrqed
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SwMarc said:
OK I am back. It turns out this problem was on the test with one variation the radius went to 15cm. I used faradays law but got the answer wrong here is what I did.

emf=(pie(.15^2)/1.5=.0471 volts

and then to find the current I=V/R=.0471/2=.0236
The flux is the area times the B field (times a cosine but here the cos gives 1) . So [itex] {d \phi \over dt} = A {\Delta B \over \Delta t} [/itex]. Looks like you forgot the change of B field, [itex] \Delta B [/itex]

Patrick
 
  • #8
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Crap, so the change of B would be two so if I put
emf=(2pie(.15^2)/1.5=.942 volts
and then current would be .942/2=.0471 ohms

Does the answer look good now?
 
  • #9
nrqed
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SwMarc said:
Crap, so the change of B would be two so if I put
emf=(2pie(.15^2)/1.5=.942 volts
and then current would be .942/2=.0471 ohms

Does the answer look good now?
I get 10 times smaller than your results ( I used 2 T* pi (0.15 m)^2/1.5 s also and got 0.0942 T m^2/s)
 
  • #10
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nrqed said:
I get 10 times smaller than your results ( I used 2 T* pi (0.15 m)^2/1.5 s also and got 0.0942 T m^2/s)
I get the same thing as you...apparently I wasnt paying atttention before.
 

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