A square loop of side 20, {cm} and resistance | vedclass

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(B) The motional electromotive force $(EMF)$ induced in the moving arm is given by $\varepsilon = B \ell {v}_{0}$.Given: $B = 5\, {T}$,$\ell = 20\, {c

Vedclass · Accepted Answer

$1\, {cm/s}$

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(B) The motional electromotive force $(EMF)$ induced in the moving arm is given by $\varepsilon = B \ell {v}_{0}$.Given: $B = 5\, {T}$,$\ell = 20\, {cm} = 0.2\, {m}$,and the internal resistance of the loop $r = 1\, \Omega$.The external circuit consists of four resistors of $4\, \Omega$ each. Looking at the circuit,the two resistors on the left are in series $(4+4=8\, \Omega)$,and the two on the right are in series $(4+4=8\, \Omega)$. These two branches are in parallel,so the equivalent external resistance $R_{eq} = \frac{8 	imes 8}{8+8} = 4\, \Omega$.The total resistance of the circuit is $R_{total} = R_{eq} + r = 4\, \Omega + 1\, \Omega = 5\, \Omega$.The current $i$ is given by $i = \frac{\varepsilon}{R_{total}} = \frac{B \ell {v}_{0}}{5}$.Given $i = 2\, {mA} = 2 	imes 10^{-3}\, {A}$,we have:$2 	imes 10^{-3} = \frac{5 	imes 0.2 	imes {v}_{0}}{5}$$2 	imes 10^{-3} = 0.2 	imes {v}_{0}$${v}_{0} = \frac{2 	imes 10^{-3}}{0.2} = 10^{-2}\, {m/s} = 1\, {cm/s}$.

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