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In figure below, a conducting rod is pulled to the right with a speed v when a f

ID: 1440571 • Letter: I

Question

In figure below, a conducting rod is pulled to the right with a speed v when a force F_app is applied. The load resistance R in the circuit is 6.0 ohm (ignoring small internal resistance of the rod and rails) and the rail separation l is 1.20 m. A uniform 2.50-T magnetic field is directed into the page, a) At what speed should the bar be moved to produce a current of 0.500 A in the resistor? b) Calculate the applied force required to move the bar to the right at a constant speed of 2.00 m/s. c) At what rate is energy delivered to the resistor in part b?

Explanation / Answer

(a)

B= 2.5 T

Induced emf = I/R =B dA/dt A is are made by loop. dA = l*dx

=0.5*6=2.5 *1.2*dx/dt

dx/dt = speed = 1 m/s.

(b)

dx/dt = 2m/s

Induced emf = IR =B*ldx/dt

I = 2.5*1.2*2 /6= 1 A

Force applied = lorrents force= I*l*B = 1*1.2*2.5 = 3 N

(c)

Energy direverd = V*I = I^2 * R =1^2 * 6 = 6 W.