The circuit shown above is located in a magnetic field B that is directed inside

Question

The circuit shown above is located in a magnetic field B that is directed inside the Figure.
The rod with mass m is sliding with velocity V on the contacts [the distance between contacts is L].
The other parameters of the circuit are shown in the Figure (E is the emf of the battery, R1 and R2 are the resistors).

1) Find acceleration of the rod, and currents I =? , I1 =?, and I2 =?.
2) If the magnetic field with the same magnitude B would be directed outside the Figure, what would be the currents I =? , I1 =?, and I2 =?, and acceleration of the rod?

Explanation / Answer

Given, magnetic field is into the paper
mass of rod = m
velocity of rod, = V
distance between contscts = L = length of the rod which has current
EMF of battery = E
resistor values = R1, R2

1) at time t
The given currenets give us
I = I1 + I2 ( kirchoff's junction law)
and from kirchoff's voltage law
E = I*R1 ( as the resistoance of the rod is 0)
and in the other loop
I2*R2 = 0
so, I2 = 0
I = I1 = E/R1
these results are while ignoring the effects of moving rod
due to moving rod, an induced current in opposite direction will be there
so, in the left loop, induced emf = V
so, V = rate of change of flux = B*dA/dt = B*Lv
so I = E/1 - BLv/R1 = (E - BLv)/R1
and I1 = -Blv/R1
similiarly
I2 = Blv/R2

now forcxe on the rod = BILsin(90) = BIL
but force = m*a ( a is acceleration of the rod)
so, BIL = ma
a = BIL/m = BL(E - BLv)/mR1
2) if the field is directed outside
I = E/R1 + BLv/R1
I1 = BLv/R1
I2 = -BLv/R2
a = BL(E + BLv)/mR1

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