Induced Current Due to a Solenoid

  • Thread starter zachfoltz
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Homework Statement


An aluminum ring of radius 5.00 cm and resistance 3.00 × 10–4 Ω is placed on top of a long air-core solenoid with 1 000 turns per meter and radius 3.00 cm. Over the area of the end of the solenoid, assume that the axial component of the field produced by the solenoid is half as strong as at the center of the solenoid. Assume the solenoid produces negligible field outside its cross sectional area. The current in the solenoid is increasing at a rate of 270 A/s. (a) What is the induced current in the ring?


Homework Equations


I = abs(emf)/R
emf = -Nd[itex]\phi[/itex]b/dt
Bsolenoid=[itex]\mu[/itex]0nI


The Attempt at a Solution


I'm somewhat confidant I know how to solve this problem but my solutions manual is telling me a different answer and the solutions manual's explanation seems wrong. What I did was simply combined all of the equations above to get:
Iring = ([itex]\mu[/itex]0nsolenoidAring/2R) (dI/dt)

The solutions manual says it should be:
Iring = ([itex]\mu[/itex]0nsolenoidAsolenoid/2R) (dI/dt)

So basically my question is which area should i use to find the flux of the magnetic field through the ring.
 

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Answers and Replies

  • #2
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If the book is right why would you use the area of the solenoid to calculate the flux?
 

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