Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same speed 5 cm s–1. Find the electric current in the 19 Ω resistor if

(a) both the wires move towards right and


(b) If P1Q2 moves towards left but P2Q2 moves towards right.



Given:


Velocity with which the wires move(v) = 5 cm s-1 = 0.05 m s-1


Resistance(R) = 19 Ω


Resistance of each of the wires(r) = 2 Ω


Length of wire(l) = 4 cm = 0.04 m


Magnetic field(B) = 1 T


Formula used:


Emf induced… (i) where B = magnetic field, l = length of wire, v = velocity


Also, by Ohm’s law, emf … (ii), where I = current, R = total resistance.


(a) When the wires slide in the same direction, we have two parallel sources of emf with current flowing in the same direction.


Hence, from (i),net emf


Net parallel resistance of the 2 Ω wires = Ω = 1 Ω


Hence, total resistance(R) = (1 + 19) Ω = 20 Ω


Therefore, substituting these values in (ii), we get


Current A = 10-4 A (Ans)


(b) When the wires slide in opposite directions, the two parallel sources of emf have opposing directions. Hence, the net emf is 0.


Therefore, the net current is also 0. (Ans)


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