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 4.4 m/s perpendicular to a 0.48-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 1.1 m. A 0.73-? 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

R = resistance = 0.73 ohm

L = length = 1.1 m

V = speed = 4.4 m/s

B = magnetic field = 0.48 T

a)

induced Voltage is given as ::

E = BL V = 0.48 x 4.4 x 1.1 = 2.32 volts

current flowing is given using ohm's law as

i = E/R = 2.32/0.73 = 3.18 A

magnetic force acting on the rod is given as::

F = I BL

the magnetic force balances the weight

hence , F = mg

mg = iBL

m = IBL/g

m = (3.18)(0.48) (1.1) /(9.8)

m = 0.171 kg

b)

distance travelled in t = 0.2 sec

distance = h = speed x t = 4.4 x 0.2 = 0.88 m

potential energy is given as ::

PE = mgh

PE = (0.171) (9.8) (0.88)

PE = 1.475 J

c)

electrical energy dissipated is given as ::

energy = i2 R t = (3.18)2 (0.73) (0.2) = 1.476 J

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