A 10 text{ cm} long perfectly conducting wire | vedclass

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Question

(A) The motional electromotive force (emf) induced in the wire $PQ$ is given by $\varepsilon = Blv$.Given: $B = 1 	ext{ T}$,$l = 10 	ext{ cm} = 0.1

Vedclass · Accepted Answer

$0.63$

Vedclass · Answer

(A) The motional electromotive force (emf) induced in the wire $PQ$ is given by $\varepsilon = Blv$.Given: $B = 1 	ext{ T}$,$l = 10 	ext{ cm} = 0.1 	ext{ m}$,$v = 1 	ext{ cm/s} = 0.01 	ext{ m/s}$.$\varepsilon = 1 	imes 0.1 	imes 0.01 = 10^{-3} 	ext{ V}$.When the key $S$ is closed,the circuit acts as an $LR$ series circuit with a constant voltage source $\varepsilon$. The current $i$ at time $t$ is given by $i(t) = \frac{\varepsilon}{R} (1 - e^{-Rt/L})$.Given: $R = 1 \ \Omega$,$L = 1 	ext{ mH} = 10^{-3} 	ext{ H}$,$t = 1 	ext{ ms} = 10^{-3} 	ext{ s}$.Substituting the values:$i = \frac{10^{-3}}{1} (1 - e^{-(1 	imes 10^{-3}) / 10^{-3}})$$i = 10^{-3} (1 - e^{-1})$Using $e^{-1} = 0.37$:$i = 10^{-3} (1 - 0.37) = 10^{-3} (0.63) = 0.63 	imes 10^{-3} 	ext{ A}$.Comparing this with $x 	imes 10^{-3} 	ext{ A}$,we get $x = 0.63$.

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