A 14.0-g conducting rod of length 1.30 m is free to slide downward between two v

Question

A 14.0-g conducting rod of length 1.30 m is free to slide downward between two vertical rails without friction. The rails are connected to an 8.00 ? resistor, and the entire apparatus is placed in a 0.480 T uniform magnetic field. Ignore the resistance of the rod and rails.

(a) What is the terminal velocity of the rod?
___m/s

(b) At this terminal velocity, calculate the rate of change in gravitational potential energy (include sign) and the power dissipated in the resistor.

Explanation / Answer

Note. V = voltage, v = velocity. Value of g = 9.81 m/s^2.
A. At terminal velocity, gravitational force equals the force on a current-carrying conductor in a magnetic field, where the current is due to voltage produced by motion of the wire in the field (motional EMF). See the refs.
V = vLB
F = ILB = mg ==> I = mg/(LB); substituting below,
v = V/(LB) = IR/(LB) = mg/(LB)*R/(LB) = 2.82174 m/s
B. Rate of change of GPE = mgv = 0.014*9.81*-2.82174 = -0.3875377 W
V = vLB = 2.82174*1.3*0.48 = 1.76076 V
P(resistor) = V^2/R = 0.38753772 W

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