A conducting rod with a weight of 2.00 N and a length of 3.00 m can slide with n

Question

A conducting rod with a weight of 2.00 N and a length of 3.00 m can slide with no friction down a pair of vertical conducting rails, as shown in the figure above. The rails are joined at the bottom by a light bulb of resistance 2.00 ohms. The rails have stops near the bottom to prevent the rod from smashing the bulb. There is a uniform magnetic field of magnitude 2.00 T directed out of the page. When the rod is released from rest, the force of gravity causes the rod to accelerate down the rails, but the rod eventually reaches a terminal velocity (that is, it falls at constant speed).

What is the magnitude of the maximum current in this situation?

Thanks

Explanation / Answer

induced voltage = Blv

Induced current = voltage / R = BLv/R

magnetic force (upward) = L i B

which must equal the weight of the rod so

W = L (BLv/R) B

Or...

v = WR / B^2 L^2 = 2.00 * 3.00 / 2.00^2 * 2.00^2 =

= 2 * 3 / 4*4 = 0.375 m/s

So the current is

i = BLv/R = 2.00 * 3.00 * 0.375 / 2.00 = 1.125 amps

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