Question

A conducting rod of mass 'm' and length 'l' is placed on a pair of smooth vertical conducting parallel rails connected at one end by an inductor of inductance L and by a capacitor of capacitance 'C' at the other end as shown in the figure. A uniform horizontal magnetic field 'B' exists in the region perpendicular to the plane of rails. The conductor starts from rest at t = 0 from the top end of the rails. [Take $B^2l^2C = 2m$].

Answer 1

The rod undergoes simple harmonic motion with angular frequency  

$\omega = \sqrt{\frac{B^2 l^2}{mC}} = \frac{B\,l}{\sqrt{mC}},$
and the maximum speed attained is
$v_{\max} = \frac{g}{\omega} = \frac{g\,\sqrt{mC}}{B\,l}.$

Answer 2

1. Circuit and motion equations

• Induced emf in rod: $\varepsilon = B\,l\,v$.
• Kirchhoff's loop:

$$L\,\frac{d i}{dt} + \frac{q}{C} = B\,l\,v,$$

with $i = \frac{dq}{dt}$.

• Newton’s law for rod:

$$m\,\frac{dv}{dt} = m\,g - B\,l\,i.$$

2. Derivation of oscillation
Differentiate Newton’s law:

$$m\,\frac{d^2 v}{dt^2} = -B\,l\,\frac{di}{dt}.$$

From the loop equation:

$$\frac{di}{dt} = \frac{1}{L}\Bigl(B\,l\,v - \frac{q}{C}\Bigr).$$

But $q = \int i\,dt$, and $v = \frac{dq}{dt}$. Combining and eliminating $i$ and $q$ yields:

$$\frac{d^2 v}{dt^2} + \frac{B^2 l^2}{m\,C}\,v = 0,$$

showing simple harmonic motion with

$$\omega^2 = \frac{B^2 l^2}{m\,C}.$$

3. Using given relation
Given $B^2 l^2 C = 2m$, one finds

$$\omega = \sqrt{\frac{B^2 l^2}{mC}} \quad\text{and}\quad v_{\max} = \frac{g}{\omega} = \frac{g\,\sqrt{mC}}{B\,l}.$$