The system shown in Figure below is in the homogeneous magnetic field B = 0.1 T.

Question

The system shown in Figure below is in the homogeneous magnetic field B = 0.1 T. The distance between two conducting vertical rails is l = 10 cm. A rod having a mass M = 10 g start moving in a gravitational field (initial velocity is zero), and it is always in electrical contact with the rails. The emf of the battery is E = 12 V, and the resistance of the resistor is R = 4 Om, Two forces are exerted on the rod: one is the gravity force (~F = m~g), and the second is the force due to the magnetic field (FA = IBl, its direction is given by the corresponding rule). At the beginning (v = 0), the current in the circuit is determined by the emf of the battery only. Under the action of these two forces, the rod starts moving, and once it acquires some velocity, the induction emf appears and changes the current in the circuit and the magnitude of the magnetic force. Eventually, the velocity of the rod reaches some magnitude at which the gravity force is equal to the magnetic force and, after that, there is no velocity change any more (because the acceleration of the rod is zero). Please, find this velocity of the rod. Find the magnitude of this velocity in the case when emf of the battery E = 0 V. What is the magnitude of this velocity if also the resistance R is zero?

Explanation / Answer

I cant see the picture, so I'll just have to guess at what it looks like. There are only two possibilities:

1. The rails are inclined, and the rod slides down the inclined rails

or

2. The rails are vertical and the rod slides straight down (vertically.)

So... I'm going to GUESS #2, that the rails are vertical. If this is wrong, there's not much I can do... because you didnt provide the picture (and you didn't bother to describe it.)

So...

upward force on total current = L I B where I is total current

total current = total V / R where R is resistance

total V = V from induction - V from battery

V from induction = B L v where v is speed

Put all this together and you get

F = LIB = L(total V / R) B = L (BLv - V) B / R

or

F = (BL/R) (BL v - V) where V is that of the battery

To make acc = 0, this force must be balanced by the force of gravity so

mg = (BL/R) (BL v - V)

or

Rmg / BL = BL v + V

so

v = Rmg/(BL)^2 - V/BL =

= 4 * 0.010 * 9.80 / (0.1*0.1)^2 + 12/0.1*0.1 =

= 3920 + 1200 = 5120 m/s

If there was no battery, then V would be zero and the calculation would give just 3920 m/s

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