Question

As shown in figure, the two parallel conducting rails, in a horizontal plane, are connected at one end by an inductor of inductance $L$ . A slider (metallic) of mass $m$ , is imparted a velocity $v_0$ _, upon the rails, as shown in figure. The period of oscillation of the conducting rod is

• A. $\frac{\pi }{2}\frac{\sqrt{mL}}{Bl}$
• B. $\frac{2\pi \sqrt{mL}}{3Bl}$
• C. $\frac{2\pi \sqrt{mL}}{Bl}$
• D. $\frac{3\pi \sqrt{mL}}{2Bl}$

Answer 1

Answer: (C)

From energy conservation (there are no dissipative forces)-
$E=\frac{1}{2}m{v}^2+\frac{1}{2}L{I}^2$
$\Rightarrow \frac{m}{2}\times 2v\frac{dv}{dt}+\frac{L}{2}\times 2I\frac{dI}{dt}=0$
$mv\frac{dv}{dt}=-IL\frac{dI}{dt}$
The emf induced in an inductor is
$E=L\frac{dI}{dt}=Blv$
$\Rightarrow L\frac{dI}{dt}=Bl\frac{dx}{dt}\Rightarrow LdI=Bldx\Rightarrow I=\frac{Bl}{L}x$
The force acting on a current-carrying conductor is
$m\frac{dv}{dt}=-IBl$
$\Rightarrow \frac{dv}{dt}=-\frac{IBl}{m}$
Substituting the value of $I$ we get
$\frac{dv}{dt}=-\left(\frac{Bl}{L}x\right)\frac{Bl}{m}\Rightarrow a=-\left(\frac{B^2{l}^2}{mL}\right)x$
We got acceleration in the form of
$a=-{\omega }^{2}x$
this implies that the motion is SHM, with angular frequency
$\omega =\sqrt{\frac{B^2{l}^2}{mL}}$
The time period of motion is
$\Rightarrow T=2\pi \frac{\sqrt{mL}}{Bl}$