A bar of mass m,length d,and resistance R sli | vedclass

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(C) The motional emf induced in the bar moving with velocity $v$ is $\varepsilon_{	ext{ind}} = Bvd$.The net emf in the circuit is $\varepsilon_{	ext

Vedclass · Accepted Answer

$v = \frac{\varepsilon}{Bd}\left(1 - e^{-\frac{B^2d^2t}{mR}}\right)$

Vedclass · Answer

(C) The motional emf induced in the bar moving with velocity $v$ is $\varepsilon_{	ext{ind}} = Bvd$.The net emf in the circuit is $\varepsilon_{	ext{net}} = \varepsilon - Bvd$.The current in the circuit is $I = \frac{\varepsilon - Bvd}{R}$.The magnetic force on the bar is $F = IdB = \left(\frac{\varepsilon - Bvd}{R}ight)Bd$.Using Newton's second law,$m \frac{dv}{dt} = \frac{(\varepsilon - Bvd)Bd}{R}$.Rearranging the terms,$\frac{dv}{\varepsilon - Bvd} = \frac{B^2d^2}{mR} dt$.Integrating both sides from $0$ to $v$ and $0$ to $t$:$\int_0^v \frac{dv}{\varepsilon - Bvd} = \int_0^t \frac{B^2d^2}{mR} dt$.$-\frac{1}{Bd} \ln\left(\frac{\varepsilon - Bvd}{\varepsilon}ight) = \frac{B^2d^2t}{mR}$.$\ln\left(1 - \frac{Bvd}{\varepsilon}ight) = -\frac{B^2d^2t}{mR}$.$1 - \frac{Bvd}{\varepsilon} = e^{-\frac{B^2d^2t}{mR}}$.$v = \frac{\varepsilon}{Bd}\left(1 - e^{-\frac{B^2d^2t}{mR}}ight)$.

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