A 13.0-g conducting rod of length 1.30 m is free to slide downward between two v

Question

A 13.0-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 0.570 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

a)

v = mgR/B^2l^2 = 0.013*9.8*8/(0.57^2*1.3^2) = 1.856 m/s <-----------answer

b)

rate change in gravitational P.E = B^2l^2v^2/R = 0.57^2*1.3^2*1.856^2/8 = -0.236 W

power dissipated n resistor = 0.236 W

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