A conducting bar moves along frictionless conducting rails connected to a 4.00 omega resistor. The length of the bar is 1.60m and a uniform magnetic field of 2.20T is applied perpendicular to the paper pointing outward as shown
a) What is the applied force required to move the bar to the right with a constant speed of 6.00 m/s?
b) At what rate is energy dissipated in the 4.00 ohm resistor?
Homework Answers
E= Blv
=2.20 x 1.6 x 6
=21.12V
now I = E/R = 21.12/4 = 5.28A
a. force on the rod = IlB = 5.28 x 1.6 x 2.2 = 18.58 N
b. rate of dissipation of energy = I^2R =5.28 x 5.28 x 4 =111.5 W
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