A 12.0-g conducting rod of length 1.30 m is free to slide downward between two v

Question

A 12.0-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 0.380 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.

__W (rate of change in gravitational potential energy)

__W (power dissipated in resistor)

Explanation / Answer

sol 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). V = vLB F = ILB = mg ==> I = mg/(LB); substituting below, v = V/(LB) = IR/(LB) = mg/(LB)*R/(LB) = 2.0608522 m/s so B. Rate of change of GPE = mgv = 0.012*9.81*-2.0608522 = -0.24260352 W we have V = vLB = 2.0608522*1.3*0.52 = 1.3931361 V P(resistor) = V^2/R = 0.24260352 W ans

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