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 4.7 m/s perpendicular to a 0.51-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 0.59- 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 0.15 s. (c) Find the electrical energy dissipated in the resistor in 0.15 s.

Explanation / Answer

a)

F_grav =F_mag

mg =BIL

since I=Einduced/R =BLV/R

mg =B^2L^2V/R

m =B^2L^2V/Rg

m =(0.51)^2*(1.1)^2*4.7/0.59*9.8

m=0.256 Kg

b)

dU =mgh =-mg(V*dt)=-0.256*9.8*4.7*0.15

dU= -1.77 J

c)

P=E^2/R =(BLV)^2/R =(0.51*1.1*4.7)^2/0.59

P=11.78 W

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