The figure below shows a rod of length L = 10.0 cm that is forced to move at con

Question

The figure below shows a rod of length L = 10.0 cm that is forced to move at constant v = 5.00m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.400. The rest of the loop has negligible resistance. A current, i = 100A, is passing through the long straight wire shown above the loop. The wire is a distance a = 10.0 mm from the loop and creates the magnetic field shown below.

Find the:
(a) emf
(b) the current induced in the loop.
(c) At what rate is thermal energy generated in the rod?
(d) What is the magnitude of the force that must be exerted on the rod to force it to move at a constant speed?
(e) At what rate does this force do work on the rod

Explanation / Answer

B(r)=µ0*I/2pr first lets find the flux f=A*B, B is changing with r. So we will break the area into small strips with width dr and length x. dA=x*dr df=B(r)*dA = (µ0*I/2pr)*x*r, then lets integrate it from a to L+a --> f=(µ0*I/2p)*x*ln((L+a)/a) then, form faraday law e=-df/dt=(µ0*I/2p)*ln((L+a)/a)*dx/dt dx/dt=v ---> e=(µ0*I/2p)*ln((L+a)/a)*v=2.4e-4 V (2) i=e/R=2.38e-4/0.4=6e-4 A (3) P=I2R=1.44e-7 Watt (4) to get the force we need to integrate dF=i*dr*B =i*dr*µ0*I/2pr = (µ0i*I/2p)*dr/r (pay attention there are two currents here i and I, one is in the bar, second is in the wire) F=integral(dF,dr)=(µ0i*I/2p)*ln((L+a)/a)=2.88e-8 N (5) this will be the same as we calculated at (3) P=F*v=2.88e-8*5=1.44e-7

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