Review Conceptual Example 3 and the drawing as an aid in solving this problem. A

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 5.3 m/s perpendicular to a 0.62-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 1.7 m. A 0.64- 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 0.21 s. (c) Find the electrical energy dissipated in the resistor in 0.21 s.

Explanation / Answer

F=bvl mg=bvl m=.62*5.3*1.7/9.8 =0.57kg in time 0.21 sec, d=.21*5.3 =1.113 therefore gravitational energy=mgd =0.57*9.8*1.113 =6.22J the electrical energy dissipated=6.22J

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