Consider the situation shown in the figure. T | vedclass

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Question

(A) The motional $emf$ induced in each wire is given by $V = B \ell v$. Given $B = 0.5\, T$,$\ell = 4\, cm = 0.04\, m$,and $v = 5\, cm/s = 0.05\, m/s$

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$0.1\, mA, 0$

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(A) The motional $emf$ induced in each wire is given by $V = B \ell v$. Given $B = 0.5\, T$,$\ell = 4\, cm = 0.04\, m$,and $v = 5\, cm/s = 0.05\, m/s$. $V = 0.5 	imes 0.04 	imes 0.05 = 10^{-3}\, V = 1\, mV$. Case $(a)$: Both wires move towards the right. Both wires act as batteries of $emf$ $V$ in parallel with internal resistance $2\,\Omega$ each. The equivalent $emf$ $E_{eff} = V = 1\, mV$ and equivalent internal resistance $r_{eff} = (2\,\Omega || 2\,\Omega) = 1\,\Omega$. The current in the $9\,\Omega$ resistor is $i = \frac{E_{eff}}{R + r_{eff}} = \frac{10^{-3}}{9 + 1} = \frac{10^{-3}}{10} = 10^{-4}\, A = 0.1\, mA$. Case $(b)$: $P_1Q_1$ moves left and $P_2Q_2$ moves right. The induced $emf$s in the two wires will be in opposite directions relative to the circuit loop. Thus,the net $emf$ in the circuit is $V - V = 0$. Therefore,the current in the $9\,\Omega$ resistor is $i = 0$.

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