Question

Two long conducting rods suspended by means of two insulating threads as shown in figure are connected at one end to a charged capacitor through a switch $S$, which initially open. At the other end, they are connected by a loose wire. The capacitor has charge $Q$ and mass per unit length of the rod is $\lambda$. The effective resistance of the circuit after closing the switch is $R$. If the velocity of each rod when the capacitor is discharged after closing the switch is $v=\frac{{\mu }_0{Q}^2}{x\pi dRC\lambda l}$. Find x.(Assume that the displacement of rods during the discharging time is negligible)

Answer 1

$F_{mag}=ma$
$\frac{{\mu }_0{I}^2}{2\pi d}=\lambda la$ (where $l=$ length of each rod)
$\frac{{\mu }_0}{2\pi d}{\left(\frac{Q}{RC}{e}^{\frac{-t}{RC}}\right)}^2=\lambda l\frac{dv}{dt}$
$\frac{{\mu }_0}{2\pi d}\times \frac{Q^2}{{\left(RC\right)}^2}\times {\int }_0^{\infty }{e}^{\frac{-2t}{RC}}=\lambda l{\int }_0^{v}dv$
$\frac{{\mu }_0{Q}^2}{2\pi dRC\cdot 2}=\lambda lv\Rightarrow v=\frac{{\mu }_0{Q}^2}{4\pi dRC\lambda l}$