Question

A straight conducting bar of mass m and length l is suspended horizontally with two non conducting springs of stiffness k as shown in figure. The capacitor is initially charged to the potential difference U. At time t = 0 the switch S is closed and the capacitor discharges. The bar start oscillating in vertical plane. Find the amplitude of these oscillation (Assume the time of discharge of capacitor is much smaller than the period T of the mechanical oscillation of the bar –

• A. $\frac{B\mathcal{l}CU}{\sqrt{2km}}$
• B. $\frac{B\mathcal{l}CU}{\sqrt{km}}$
• C. $\frac{B\mathcal{l}CU}{\sqrt{8km}}$
• D. $\frac{B\mathcal{l}CU}{4\sqrt{km}}$

Answer 1

Answer: (A)

$\frac{B\mathcal{l}CU}{\sqrt{2km}}$

Answer 2

As capacitor discharge in time Dt, a current will flow due to which magnetic force will act on bar for small time. Impulse of force = F Dt = i l B Dt = $\frac{\Delta q}{\Delta t}$ l B Dt

= B l Dq Dq is the total charge flowing through capacitor in time Dt. Change in momentum = mv

mv = B l Dq mv = B l CU …(1) [Dq was the charge in C ie. CU] two spring are in parallel \ equivalent stiffness

constant = 2k

energy conservation $\frac{1}{2}mv^{2}$= $\frac{1}{2}$ × 2 k A² where A is amplitude A =$\frac{B\mathcal{l}CU}{\sqrt{2km}}$