The figure shows a rod of length L = 9.45 cm that is forced to move at constant

Question

The figure shows a rod of length L = 9.45 cm that is forced to move at constant speed v = 7.02 m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.479 ohm; the rest of the loop has negligible resistance. A current i = 150 A through the long straight wire at distance a = 10.2 mm from the loop sets up a (nonuniform) magnetic field throughout loop. Find the (a) magnitude of the emf and (b) 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 applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod?

Explanation / Answer

part a )

Letting x be the distance from the right end of the rails to the rod, we find an expression for the magnetic flux through the area enclosed by the rod and rails.

B = uoI/2pir

We consider an infinitesimal horizontal strip of length x and width dr, parallel to the wire and a distance r from it; it

has area A = x dr and the flux is

dphi = BdA = uo*I/2pir * xdr

phi = uoi*x/2pi integral dr.r from a to a+L

phi = uo*i*x/2pi ln(a+L/a)

emf = dphi/dt

e = (uoi/2pi) * (dx/dt) * ln(a+L/a) = uo*iv/2pi ln(a+l/a)

e = 2 x 10^-7 x 150 x 7.02 x ln( 1.02 cm + 9.45cm / 1.02 cm)

e = 4.9 x 10^-4 V

part b )

induced current = e/R = (4.90 x 10^-4) / (0.479) = 1.023 x 10^-3 A

part c )

P = induced current^2 x R

P = 5.01 x 10^-7 W

part d )

Since the rod moves with constant velocity, the net force on it is zero. The force of the external agent must have the

same magnitude as the magnetic force and must be in the opposite direction.

dFb = induced current*Bdr = uo*i*I/2pir * dr

Fb = uo*i*I/2pi integrate dr/r from a to a+L

Fb = uo*i*I/2pi * ln( a+L/a)

Fb = 2x10^-7 x 1.023 x 10^-3 x 150 ln( 1.02 cm + 9.45 cm / 1.02 cm )

Fb = 7.15 x 10^-8 N

part e )

P = Fv

P = 7.15 x 10^-8 x 7.02

P = 5.0193 x 10^-7 W

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