I need part B online A conducting rod slides down between two frictionless verti

Question

I need part B online

A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 5.1 m/s perpendicular to a 0.40-T magnetic field. The resistance of the rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 0.99 m. A 0.68- resistor is attached between the tops of the tracks. (a) What is the mass of the rod? (b) Find the change in the gravitational potential energy that occurs in a time of 0.28 s. (c) Find the electrical energy dissipated in the resistor in 0.28 s. (a) Number .12 Units kg (b) Number Units J (c) Number 1.68 Units J

i have seen tried all the cramster strategies. part b is wrong for all of them. i got .45 for part B using their strategies and it is wrong. please help.

Explanation / Answer

the length does change, at a rate of 5.1 m/s.

So the rate of change of flux is (0.4T)(0.99m)(5.1m/s) = 2.0196 volts,

By Ohm's law, V = IR,

we have 2.0196 V = (0.68 ohm) I and I = 2.97 amps.

mg = ILB

m = ILB/g

(2.97 amp)(0.99m)(0.4T)/(9.8m/sec²) = m = 0.120 kg.

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