A coil of self-inductance L is connected at o | vedclass

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(A) As the connector moves with velocity $v$,the motional $EMF$ induced is $\varepsilon = Blv$. This $EMF$ drives a current $I$ through the inductor $

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Vedclass · Answer

(A) As the connector moves with velocity $v$,the motional $EMF$ induced is $\varepsilon = Blv$. This $EMF$ drives a current $I$ through the inductor $L$,given by $\varepsilon = L \frac{dI}{dt}$,so $L \frac{dI}{dt} = Blv$. The magnetic force on the connector is $F = -IlB$,which causes deceleration: $m \frac{dv}{dt} = -IlB$. Substituting $I$ from the first equation,we get $m \frac{dv}{dt} = -\frac{B^2 l^2}{L} \int v dt$. Differentiating with respect to $t$,we get $m \frac{d^2v}{dt^2} = -\frac{B^2 l^2}{L} v$. This is the equation of simple harmonic motion,but since the velocity $v$ decreases and the current builds up,the connector eventually stops. The displacement $x(t)$ is given by $x(t) = \frac{mv_0}{\omega L} (1 - \cos(\omega t))$ where $\omega$ is related to the circuit parameters. However,in this specific physical setup,the connector moves and eventually comes to a stop at a maximum displacement. The graph of displacement $x$ versus time $t$ starts from the origin,increases,and approaches a constant value as $t 	o \infty$.

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