Question

A conducting rod PQ of mass ‘m’ and of length ‘l’ is placed on two long parallel(smooth and conducting) rails connected to a capacitor as shown below. The rod PQ is connected to a non-conducting spring constant ‘k’, which is initially in relaxed state. The entire arrangement is placed in a magnetic field perpendicular to the plane of figure.
Neglect the resistance of rails and rod. Now, the rod is imparted a velocity v₀ towards right, then acceleration of the rod as a function of its displacement ‘x’ is given by.

• A. $\frac{kx}{m}$
• B. $\frac{kx}{m+B^2{l}^{2}c}$
• C. $\frac{kx}{m-B^2{l}^{2}c}$
• D. None of these

Answer 1

The correct option is C $\frac{kx}{m-B^2{l}^{2}c}$

Let the velocity of rod be $v_0$ when it has been displaced by ‘x’ due to motion of rod an emf, will be induced in rod given by e = $BL{v}_0$, due to this induced emf, charging of the capacitor takes place as a current, flows in the circuit [for very small time] as a result of this current, the rod experiences a magnetic force given by IBL.



From Newton's 2nd law,
$LIB+Kx=ma$
$\Rightarrow I=\frac{d}{dt}\left(Q\right)=\frac{d}{dt}(C\times BvL)=CBL\times \frac{dv}{dt}$
$\Rightarrow a=\frac{Kx}{m-B^2{L}^{2}C}=w^{2}x$