As shown in the illustration below, a conducting bar is pulled to the right alon
ID: 1493683 • Letter: A
Question
As shown in the illustration below, a conducting bar is pulled to the right along frictionless, parallel conducting rails (l = 0.10 m apart) at a constant speed v. The end of the rails are connected by a resistor R = 3.0 Ohm. (We can assume the resistances of the bar and the rails are negligible.) Since the bar is moving through a region with a magnetic field B = 0.60 T, pointing into the page, a current I = 0.50 A is induced in the circuit loop formed by the bar, rails, and resistor. (Assume the bar is in good electrical contact with the rails.) What is the direction of this induced current in the moving bar? Find the speed v at which the bar is being pulled to induce this current. Determine the pulling force needed to move the bar at this constant speed. Find the mechanical power delivered by the force found in part (c) to move the bar at a constant speed. (Note that power is the rate that work is done, i.e., P = dW/dt = F rightarrow. dx rightarrow/dt since W = F rightarrow Delta x rightarrow.) Determine the rate of Joule heating (electric power loss) in the resistor as the bar is pulled at this speed. (Show all work to receive full credit.)Explanation / Answer
Direction of cuurent is counter clockwise to oppose the increasing flux
EMF = - d( phi) /dt = Blv
Current = Blv/R
0.5 = 0.6*0.1* v / 3 So v = 25m/s
F - IBL = 0 So F = B * Blv /R * L = B^2 L^2 V /R = 0.6 *0.5*0.1 = 0.03N
Power = F *V = 0.03 * 25 = 0.75W
Joule heat = I^2 R = B^2 L ^2 V ^2 /R = 0.75 W
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