A conducting rod slides down between two frictionless vertical copper tracks at

Question

A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.2 m/s perpendicular to a 0.51-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 1.4 m. A 0.72-? resistor is attached between the tops of the tracks.

(a) What is the mass of the rod?
kg

(b) Find the change in the gravitational potential energy that occurs in a time of 0.20 s.
J

(c) Find the electrical energy dissipated in the resistor in 0.20 s.
J

Explanation / Answer

a)

Here ,

for moving with constant speed ,

magnetic force = weight of rod

B*I*L = mg

B*L * B*v*L/R = mg

0.51^2 * 1.4^2 * 4.2 / 0.72 = m * 9.8

m = 0.303 Kg

the mass of rod is 0.303 Kg

b)

change in gravitational potential energy = - m*g*h

change in gravitational potential energy = - mg * v* t

change in gravitational potential energy = - 0.303 * 9.8 * 4.2 *0.2

change in gravitational potential energy = -2.49 J

the change in gravitational potential energy is -2.49 J

c)

electrical energy dissipated = - change in gravitationa potential energy

electrical energy dissipated = - (-2.49) J

electrical energy dissipated = 2.49 J

the electrical energy dissipated is 2.49 J

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