Solenoid induced EMF problem

  • #1
A 1 m long large coil with a radius of 15.2 cm and 280 turns surrounds a 7.6 m long solenoid with a radius of 5.2 cm and 3200 turns. The current in the solenoid changes as [tex] I=I_0 sin (2\pi f t) [/tex] where I_0= 30 A and f=60 Hz.Inside solenoid has 3200 turns and outside coil has 3200 turns. The equation for the emf is [tex] E= E_0 sin \omega*t [/tex]. There is also a resistor on the smaller coil that is 32 ohms. Find the maximum induced emf in the large coil. Answer in units of V.

I really have no idea where to start this problem.. can someone point me in the right direction?
 

Answers and Replies

  • #2
Andrew Mason
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Punchlinegirl said:
A 1 m long large coil with a radius of 15.2 cm and 280 turns surrounds a 7.6 m long solenoid with a radius of 5.2 cm and 3200 turns. The current in the solenoid changes as [tex] I=I_0 sin (2\pi f t) [/tex] where I_0= 30 A and f=60 Hz.Inside solenoid has 3200 turns and outside coil has 3200 turns. The equation for the emf is [tex] E= E_0 sin \omega*t [/tex]. There is also a resistor on the smaller coil that is 32 ohms. Find the maximum induced emf in the large coil. Answer in units of V.

I really have no idea where to start this problem.. can someone point me in the right direction?
The induced emf is given by the rate of change of flux enclosed by the coil. There are 3200 coils, so the emf is 3200 times the rate of change of flux enclosed by each turn of the outer coil.

The field inside the inner solenoid (air centre) is [itex]\mu_0IN/L[/itex].

So you can work out the flux inside the inner coil. To find the flux enlosed by the outer coil, I think you just have to use the same field but the area of the outer coil.

Then you have to take the time rate of change of that flux and multiply by the number of coils.

The resistor will only influence the phase difference between E and I. Since they are only asking for maximum E I don't think you have to worry about it.

Hope that helps.

AM
 
  • #3
Since the flux= B*A, how can I find it when they don't give the magnetic field?
 
  • #4
Hootenanny
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You can calulate the flux.

(1)Calculate the flux density (B) using the equation andrew gave you above.

(2)Calculate the area of the solenoid

(3)Multiply 1 and 2 together,

~H
 
  • #5
Ok so I used [tex] \mu_0 I N/L [/tex]
[tex] (1.26e-6)(30 sin 120 \pi t)(3200)/ .052 [/tex]
and got [tex] 2.33 sin 120 \pi t [/tex] for the magnetic field, then the area of the solenoid= [tex] \pi (0.052)^2 [/tex]
and I got [tex] .0198 sin 120 \pi t[/tex] for the flux.. am I doing this right?
 
  • #6
Hootenanny
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[itex]L[/itex] is the length of the solenoid not the radius. Apart from that you calcs look fine.

~H
 
  • #7
Ok so the flux of the inner coil is [tex] 1.35e-4 sin 120 \pi t [/tex]
and the outer coil is [tex] .00115 sin 120 \pi t [/tex]
Andrew Mason said:
Then you have to take the time rate of change of that flux and multiply by the number of coils.
I'm a little confused on this part. Would I just take the derivatives of the flux of the coils with respect to t?
 
  • #8
Hootenanny
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Punchlinegirl said:
I'm a little confused on this part. Would I just take the derivatives of the flux of the coils with respect to t?

Yes, if you look at faraday's law, it is basically a differential equation;

[tex] emf = -N\frac{d\phi}{dt}[/tex]

~H
 
  • #9
sorry that I'm asking so many questions, but I don't really know what I'm doing...
so the derivative of the outside coil is [tex] .4335 cos (120 \pi t)[/tex]
and the inside is [tex] .0509 cos (120 \pi t) [/tex]
To get the difference in flux, do I subtract the inside coil from the outside?
and then multiply it by 280?
 

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