Question

a pair of parallel conducting wires lies in a plane making an angle with the horizontal plane.the upper end of these wires is connected to a capacitor of capacitance C,a uniform magnetic field B is applied in the region in a direction normal to the plane of the wires.a conducting filament of mass m slides down the parallel wires starting from rest.it is possible for the filament to acquire a steady speed?if YES find the speed.if NO find the acceleration of the filament.(ignore frictions and all resistances)

Answer 1

This is a great problem! Here's my shot at it: Let the separation of the wires be \(w\), the angle of the plane of the wires to the horizontal be \(\theta\), the charge on the capacitor to be \(Q\), and the distance of the filament from the capacitor be \(x\). I will assume the conducting wires to have no resistance, and the filament to have resistance \(R\). The magnetic flux through the capacitor-wire-filament loop is\[\Phi_B=Bwx.\]Taking the time derivative gives us the induced EMF by Faraday's law. Let the velocity of the filament be \(v\).\[\frac{d\Phi_B}{dt}=Bw\frac{dx}{dt}=Bwv \overset{\text{Faraday}}{=} |\mathcal E|\]Since charges may move freely, the voltage drop across the capacitor will be the induced EMF.\[\mathcal E = \frac{Q}{C}=Bwv\]Taking the time derivative of both sides gives the following. Let the filament's acceleration be \(a\).\[\frac{\left( \dfrac{dQ}{dt} \right)}{C}=Bw\frac{dv}{dt} \longrightarrow I=BCwa\]The magnetic force on this wire is as follows if we note that the filament is \(w\) in length. It's negative because it's in the direction of decreasing \(x\).\[F_B=-IwB\]So, the net force is as follows.\[F_{\text{net}}=mg\sin \theta-IwB=mg\sin \theta - B^2w^2 C a \]In the case of terminal velocity, we'd have to have \(F_\text{net}=0\) and \(a=0\), which restricts \(mg \sin \theta=0\), which clearly only occurs when \(\theta=0\). Thus, it must be accelerating. Newton's second law tells us how much it is accelerating.\[mg \sin \theta - B^2w^2C a=ma \longrightarrow \boxed{a=\dfrac{mg \sin \theta}{B^2 w^2 C + m}}\]