Just part B Just part B A conducting rod slides down between two frictionless ve

Question

Just part B

Just part B A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.2 m/s perpendicular to a 0.46 T magnetic field. The resistance of the rod and track is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.2 m. A 0.67-ohm resistor is attached between the tops of the tracks. (a) What is the mass of the rod? 0.195 kg (b) find the change in the gravitational potential energy that occurs in a time of 0.20 s. 0.3745 J (c) Find the electrical energy dissipated in the resistor in 0.20s. J

Explanation / Answer

change in Gravitational Potential energy

dU = m g h

here h = v dt

so

dU = 0.196* 9.81 * 4.2 * 0.2

dU = -1.61 Joules

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