Question

A wire of mass m and length l can slide freely on a pair of smooth, vertical rails (figure). 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

EMF is Equal to Product of Length, Field and Speed

Explanation:

• Let the rod has a velocity v at any instant,

Then at the point,

• e = Blv

• Now, $q = c \times potential$ = ce = CBIv

Current, I = $\frac {dq}{dt} = \frac {dCBIv}{dt}$

= $CBI\frac {dv}{dt}$= CBIa (a is acceleration)

From figure, force due to magnetic field and gravity are opposite to each other. So, mg – IlB = ma

• Hence, $mg - CBla \times lB = ma$ or $ma + CB^2l^2a$ = mg

$a(m + CB^2l^2)$ = mg

or a = $\frac {mg}{m + CB^2l^2}$