Question

A 10 cm 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 = 1 mH and a resistance R = 1 $\Omega$ as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = IT perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is $xX10^{-3}$ A, 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 ]

Answer 1

Answer: 0.62

Since velocity of PQ is constant. So emf developed across it remains constant.
$\epsilon$ = Blv where $\ell$ = length of wire PQ
Current at any time t is given by
$i=\frac{\epsilon }{R}\left(1-e{}^{-\frac{Rt}{L}}\right)$
$i=\frac{B\ell v}{R}\left(1-e{}^{\frac{Rt}{L}}\right)$
$=1\times \left(\frac{10}{100}\right)\times \left(\frac{1}{100}\right)\times \frac{1}{1}\left(1-e^{\frac{-1\times 10^{-3}}{1\times 10^{-3}}}\right)$
$=\frac{1}{1000}\times \left(1-e^{-1}\right)$
$=\frac{1}{1000}\times \left(1-0.37\right)$
$i=0.63\times 10^{-3}A\Rightarrow x=0.63$