A conducting rod of length l = 35.0 cm is free to slide on two parallel conducti

Question

A conducting rod of length l = 35.0 cm is free to slide on two parallel conducting bars as shown in figure. Two resistors R_1 = 2.00 ohm and R_2 = 5.00 ohm are connected across the ends of the bars. The rod has a resistance of R = 1.00 ohm A constant magnetic field B = 2.50 T is directed perpendicularly into the page. An external agent pulls the rod to the left with a constant speed of v = 8.00 m/s. Find a) the currents in resistors R_1 R_2 and R. b) the total power delivered to the resistance of the circuit. c) the magnitude of the applied force that is needed to move the rod with this constant velocity.

Explanation / Answer

Once we figure out that the moving rod will have an induced EMF, then it’s just a battery

with two resistors in parallel.

Think of the left side as a loop whose Flux area is decreasing, and the right side as a loop whose

Flux area is increasing. Thus the right side will try to produce a magnetic field that opposes the

increased flux (a field pointing out of the page, using the right hand rule, this tells us it will be a

counter-clockwise current), and the left side will try to augment the decreasing flux–again using the

right hand rule and pointing our thumb into the page, a clockwise current.

Call x the distance between the middle rod and the right resistor, then the distance between the

middle rod and the left resistor is L x. The time derivative of this distance is dx/dt = v for the

right, and dx/dt = v for the right one. Thus the sign takes care of the directional question.

The induced EMF will thus be (for the right; for the left flip the sign):

E = -d(phi)/dt = -Blv

where l is the length of either the left or right side and phi = BA = Bxl on the right or B(L x)l on

the left.

Because the resistors are in parallel, they are both under the voltage of | Bvl |, and with I = V /R,

I1 =Bvl/Rl = 3.5A

and

I2 = Bvl/Rr = 1.4A

Using the junction rule, that tells us that a total of 4.9A rides through the rod, and the power

supplied is thus

b) Power P = VI = Bvl(4.9A) = 34.3W

c) The magnitude of the force could be found by F = P/v = 34.3/8 = 4.29N

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