Question

Review Conceptual Example 3 and the drawing as an aid in solving this problem. A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 5.9 m/s perpendicular to a 0.52-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 1.3 m. A 0.56- 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 0.20 s. (c) Find the electrical energy dissipated in the resistor in 0.20 s.

Answer 1

a)

B = magnetic field = 0.52 T

v = speed = 5.9 m/s

L = length of rod = 1.3 m

Induced emf is given as

E = BLv

E = (0.52) (1.3) (5.9)

E = 4 Volts

R = resistance = 0.56 ohm

Current in the circuit is given using ohm's law as

i = E/R

i = 4/0.56

i = 7.14 A

using equilibrium of force

magnetic force = weight

i B L = mg

(7.14) (0.52) (1.3) = m (9.8)

m = 0.5 kg

b)

h = height gained = v t = 5.9 x 0.2 = 1.2 m

Change in gravitational potential energy is given as

U = mgh = 0.5 x 9.8 x 1.2 = 5.88 J

c)

Electrical energy dissipated is given as

Q = i²R t = (7.14)² (0.56) (0.20) = 5.71 Joules