I need help with parts C through E. I have the calculated values for RL and RR a

Question

I need help with parts C through E. I have the calculated values for RL and RR as 1.5V and .75V respectively and have determined the directions using lenz's law. I'm not sure if loop rules are applicable in this situation to solve for part C forward. Thanks for the help.

Problem 3. Two conducting rods of length 0.300 m move on parallel rails in a uniform magnetic field B = 0500 T pointing into the page. The rod on the left has a resistance RL-12.0 and the rod on the right has a resistance RR-24.0n. There is also a stationary resistor R 6.00 connecting the rails. The rails themselves have negligible resistance. The rod on the left is being pulled to the left at a constant speed of UL-10.0 m/s. The rod on the right is being pulled to the right at a constant speed of VR = 5.00 m/s. (a) What is the induced emf produced by the motion of the rod on the left? (You should justify your choice of direction for the emf using Lenz's law.) (b) What is the induced emf produced by the motion of the rod on the right? (You should justify your choice of direction for the emf using Lenz's law.) (c) What are the currents (magnitude and direction) flowing through each rod and through R? (d) What is the total power dissipated by the three resistances? (e) Show that the power supplied by the two external forces equals the power dissipated by the resistors.

Explanation / Answer

c) current through left rod is iL = VL/RL = 1.5/12 = 0.125 A downwards

current through right rod is iR = VR/RR = 0.75/24 = 0.03125 A upwards

current through R is i = 0.125-0.03125 = 0.09375 A upwards

d) total power dissipated is P = (0.125^2*12)+(0.03125^2*24)+(0.09375^2*6) = 0.2636 W

e) Power supplied is P = (1.5^2/12)+(0.75^2/24) = 0.21 W

since answers for d) and e) are almost same

hence the power supplied by the two external forces equals the power dissipated by the resistors

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