Two long conducting rods suspended by means of two insulating threads as shown in Fig. are connected to a charged capacitor through a switch which is initially open. At the other end, they are connected by a loose wire.
The capacitor has a charge $Q$ and mass per unit length of the rod is $λ$ The effective resistance of the circuit after closing the switch is $R$ .
Find the velocity of each rod when the capacitor is discharged after closing the switch. Assume that the displacement of rods during the discharging time is negligible.

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$\frac{2 μ_{0} Q^{2}}{3 π λ dRC}$

$\frac{μ_{0} Q^{2}}{π λ dRC}$

$\frac{3 μ_{0} Q^{2}}{π λ dRC}$

$\frac{μ_{0} Q^{2}}{4 π λ dRC}$

answer is D.

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Detailed Solution

The impulsive force due to (antiparallel) discharging currents is equal to the change in momentum or

$λ v = \int_{0}^{\infty} Fdt = \int_{0}^{\infty} \frac{μ_{0} i^{2}}{2 π d} d t$
where $i = d i s c h a r g i n g c u r r e n t$
Now $q = Q e^{\frac{- t}{R C}}$ $| i | = \frac{Q}{R C} e^{- t / R C}$
Magnetic force per unit length between two current carrying wires is
$\frac{F}{ℓ} = \frac{μ_{0} i_{1} i_{2}}{2 π d} = \frac{μ_{0}}{2 π d} {(\frac{Q}{R C} e^{- t / R C})}^{2}$
Impulse due to this force $= \frac{μ_{0}}{2 π d} \frac{Q^{2}}{R^{2} C^{2}} e^{- 2 t / R C} dt .$
Hence
$λ v = ∣ \int_{0}^{\infty} \frac{μ_{0}}{2 π d} \frac{Q^{2}}{R^{2} C^{2}} e^{- 2 t / R C} d t = \frac{μ_{0} Q^{2}}{4 π dRC}$
or
$v = \frac{μ_{0} Q^{2}}{4 π λ dRC}$