Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm s–1. Find the current in the 10Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.
Given:
Speed(v) = 5 cms-1 = 0.05 ms-1
External resistance(R) = 10Ω
Magnetic field(B) = 1 T
Formula used:
Induced emf … (i), where B = magnetic field, l = length of sliding wire, v = velocity
(a) When the switch S is thrown to the middle rail, length of sliding wire l = 2 cm = 0.02 m
Hence, induced emf in this case from (i) is
E = (1 x 0.02 x 0.05) V = 10-3 V
Given resistance R = 10Ω
Therefore, current flowing through the resistor
where E = emf, R = resistance.
= 10-4 A = 0.1 mA (Ans)
(b) When the switch S is thrown to the bottom rail, length of sliding wire(l’) = 4 cm = 0.04 m
Hence, induced emf = (1 x 0.04 x 0.05) V = 2 x 10-3 V, where B = magnetic field, l’ = length of sliding wire, v = velocity
Resistance R = 10Ω
Therefore, current flowing through the resistor I’ = E’/R, where E’ = emf, R = resistance
A = 2 x 10-4 A= 0.2 mA (Ans)