Eduprobe
Question:
A 10cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L=1mH and a resistance R=1Ω as shown in figure. The horizontal rail, L and R lie in the same plane with a uniform magnetic field B=1T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x×10⁻³A, where the value of x is ____ . [Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given: e⁻¹=0.37, where e is base of the natural logarithm]
0.63
Solution:
Correct option is A. 0.63
ε=(→v × →B)l = Bvl = 10⁻² × 1 × 10⁻¹ = 10⁻³volt
i = ε/R(1 - e⁻Rt/L) = 10⁻³/1(1 - e⁻¹)
i = 10⁻³(1 - 0.37)
I = 0.63mA