A conducting rod slides down between two frictionless vertical copper tracks at
ID: 2188421 • Letter: A
Question
A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.3 m/s perpendicular to a 0.55-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 0.81-? 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.
Explanation / Answer
a) equating forces, mg = BIl so, m = BIl/g = Bl*(Blv/R)/g = B^2l^2v/Rg = 0.236 kgRelated Questions
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