Q. 445.0( 1 Vote )

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.



Answer :

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)


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