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?
Emf = -Blv, I = |-Blv|/R = Blv/R. F=ILB
The Attempt at a Solution
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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