Formula for average current on a coil inside a solenoid

AI Thread Summary
The discussion focuses on calculating the average current induced in a small coil placed inside a solenoid as the current in the solenoid drops to zero. The initial magnetic flux through the inner coil is calculated to be 8.88 x 10^-6 Wb, and the average induced EMF during the 15 ms drop is determined to be 0.059 V. The direction of the induced current in the small coil is counterclockwise, as the magnetic flux decreases when the solenoid's current is interrupted. To find the average induced current, Ohm's law is suggested, utilizing the previously calculated EMF and the coil's resistance of 2.50 Ω. The discussion highlights the importance of understanding electromagnetic induction principles in solving the problem.
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Homework Statement


A solenoid of length 20.0 cm is made of 5000 circular coils. It carries a steady current of 10.0 A. Near its center is places a very flat and small coil with a resistance of 2.50 Ω made of 100 circular loops, each with a radius of 3.00 mm. This small coil is oriented so that its area receives the maximum magnetic flux. A switch is opened in the solenoid circuit and its current drops to 0 in 15.0 ms. (a)What is the initial magnetic flux through the inner coil? (b) Determine the average induced EMF in the small coil during the 15.0 ms. (c) If you look along the long axis of the solenoid so that the official 10.0 A current is clockwise, determine the direction of the induced current in the small inner coil during the time that the current drops to zero. (d) What is the magnitude of the average induced current in the coil?


Homework Equations


I don't know which equation I need!



The Attempt at a Solution


I have solved everything up until part d. I don't know which equation I need to use to find the average induced current in the inner coil.
This is what I have done for parts a through c:
a) B=μnI
B= 4∏x10^-7(5000/.200)10.0= 0.31416 F
\Phi=BAcosθ
0.31416(∏0.003^2)cos(0)= 8.88x10^-6 Wb
b) [ε]=-N(Δ\Phi/Δt)
-100(8.88x10^-6/0.015)= 0.059 V
c) CCW

Thanks so much for your help! And if you wouldn't mind checking the parts I completed I would be so grateful!
 
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For (d) use your result from (b) and the given resistance of the coil. Think Ohm's law.

You might want to reconsider your answer to (c). Note, in this problem the flux through the coil decreases when the switch is opened.
 
Haha of course! I feel silly for not thinking of that. Thanks so much TSny!
 
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