(1) A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are
(1) I1 = I2 = Blv/6R, I = Blv/3R
(2) I1 = – I2 = Blv/R, I = 2Blv/R
(3) I1 = I2 = Blv/3R, I = 2Blv/3R
The emf induced in the sliding connector PQ is ε = Blv, which is shown as a battery of internal resistance R in the adjoining figure.
The two external resistances R and R appear in parallel with this battery. The effective external resistance is therefore R/2 and the battery has to drive the current I through a total resistance R + (R/2), which is 3R/2
Therefore, I = ε/(3R/2) = 2Blv/3R.
The current I gets divided equally between the two identical branches and hence I1 = I2 = Blv/3R.
(1) V(R1 + R2)/R1R2 at t = 0 and V/R2 at t = ∞
(2) VR1R2/√(R12+ R22) at t = 0 and V/R2 at t = ∞
(3) V/R2 at t = 0 and V(R1 + R2)/R1R2 at t = ∞
(4) V/R2 at t = 0 and VR1R2/√(R12+ R22) at t = ∞
The current through the inductance branch is zero at the instant the circuit is switched on. [Remember the exponential growth of current in an LR circuit]
Therefore, at the instant t = 0 the current is limited by R2 alone and is equal to V/R2.
At t = ∞ the current has become steady, there is no induced voltage to oppose the flow of current and the inductor functions as a piece of conductor. The current is now limited by R1 and R2 which are in parallel with the battery. The current in this case is V/[R1R2/(R1 + R2)] = V(R1 + R2)/R1R2, as given in option (3).
You will find similar useful multiple choice question (with solution) here.
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