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Electromagnetic induction in a U shaped conductor

  • #1

Homework Statement


[/B]
A 0.393 m long metal bar is pulled to the left by an applied force F. The bar rides on a parallel metal rails connected through a 42.9 ohm resistor as shown in the figure. So the apparatus makes a complete circuit. You can ignore the resistance of the bars and the rails. The circuit is in an uniform 0.34 T magnetic field that is directed out of the plane of the figure. At the instant when the bar is moving to the left at 5.9 m/s. what is the rate at which the applied force is doing work on the bar?



Homework Equations


Emf = -Blv, I = |-Blv|/R = Blv/R. F=ILB

The Attempt at a Solution


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For my answer, I know that F=ILB but I=BLv/R so F=(B^2*L^2*v)/R. And we know that power = Force * velocity so we get: P = Fv = (BLV)^2/R. I subbed in the values and I got 14.5 mW. However, the text book disagrees and says its 50.4 mW, where did I go wrong?
 

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Answers and Replies

  • #2
Hesch
Gold Member
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I think your result is correct: Calculated different:

ψ = B * A , ψ is the flux , A is area.

Emf = dψ/dt = B*dA/dt = 0.34T * 0.393m * 5.9m/s.

P = Emf2 / R
 
  • #3
TSny
Homework Helper
Gold Member
12,405
2,841
The wording of the problem does not necessarily imply that the applied force equals the magnetic force (BLv). For example, the rod could have a nonzero acceleration "at the instant when the bar is moving at 5.9 m/s to the left". The sum of the applied force vector and magnetic force vector would then equal mrod ##\vec{a}## (assuming no friction between the rod and the rails).

Anyway, to get a definite answer it seems natural to make the assumption of constant speed and no friction so that the applied force equals the magnetic force. So, I think your answer is correct.
 

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