Question

A wire of mass m and length l can slide freely on a pair of smooth, vertical rails (figure 38-E31). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.

Answer 1

Let the rod has a velocity '$\theta$' at any instant, then, at that point,

$e=Bl\theta$

Now, $q=c\times potential=ce=cBl\theta$

Current$=\frac{dq}{dt}=cBl\theta$

$=cBl\frac{d\theta }{dt}=cBla$

(where $a\to$ acceleration)

From given figure force due to magnetic field and gravity are opposite to each other.

So, $mg-ilB=ma$

$\Rightarrow mg-cBla\times lB=ma$

$\Rightarrow ma+c{B}^2{l}^{2}a=mg$

$\Rightarrow a(m+c{B}^2{l}^2)=mg$

$\Rightarrow a=\frac{mg}{m+c{B}^2{l}^2}$