A conducting rod slides down between two frictionless vertical copper tracks at
ID: 1961827 • Letter: A
Question
A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.5 m/s perpendicular to a 0.56-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 1.2 m. A 0.74- resistor is attached between the tops of the tracks.
A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.5 m/s perpendicular to a 0.56-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 1.2 m. A 0.74-ohm 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. JExplanation / Answer
Given that speed v = 4.5 m/s Magnetic field B = 0.56 T length L = 1.2 m resistance R = 0.74 O -------------------------------------------------------- The magnitude of induced emf is e = BLv = (0.56 T)(1.2 m)(4.5 m/s) = 3.024V The current in the circuit is I = e/R = ( 3.024 V)/(0.74 ohm) = 4.086 A The magnetic force is F = BIL But F =mg So the mass of the rod is m = (BIL)/g = (0.56 T)(4.086 A)(1.2 m)/(9.80 m/s2) = 0.2802 kg b) From the kinematic relations s = ut+ (1/2) gt^2 = 0 +(1/2)gt^2 = (0.5)(9.80 m/s2)(0.20 s)^2 = 0.196 m The change in the gravitational potential energy is U = mgh =( 0.2802 kg)(9.80 m/s2)(0.196 m) = 0.5382 J c) the electrical energy dissipated in the resistor in 0.20 s is E = VI(t) = ( 3.024 V)( 4.086 A)(0.20 s) = 2.471 U
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