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Induced emf in coil with decreasing current

  • Thread starter kopinator
  • Start date
  • #1
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A single-turn circular loop of radius R = 0.197 m is coaxial with a long 1740 turn solenoid of radius 0.0410 m and length 0.890 m, as seen in the figure below. (picture in URL)

https://s1.lite.msu.edu/enc/53/3b1b...bbf37ae59602796c87a446f0a53c5801a82918f44.gif

The variable resistor is changed so that the solenoid current decreases linearly from 6.81 A to 1.59 A in 0.225 s. Calculate the induced emf in the circular loop. (The field just outside the solenoid is small enough to be negligible.)

phi(flux)= ∫B*dA
ε= dphi/dt
A(circle)=∏r^2
B= N*mu_0_*I/L (solenoid)

I'm having troubles finding the flux through the loop. I tried taking the integral from .0410 to .197 m but I don't think that is right. I know once I get my flux integral i can take the flux at both currents, find the difference between the two, and divide by .225 s to find the induced emf. I just don't know what to integrate over.
 

Answers and Replies

  • #2
TSny
Homework Helper
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Recall that for a long (ideal) solenoid:
1. B is essentially uniform inside the solenoid
2. B is very weak outside the solenoid so that, to a good approximation, you can assume B = 0 outside the solenoid.
 
  • #3
41
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So would i only integrate from 0 to .0410 then?
 
  • #4
TSny
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