Question

Two metal bars are fixed vertically and are connected on the top by a capacitor C. A sliding conductor AB of length L slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. The displacement (x) in meter of the conductor at time $t=2sec$ is:
(Given $m=100gm,g=10m/s^2,B=100Tesla,L=1m,andC=10\mu F$)

• A. $10$
• B. $14$
• C. $7$
• D. $5$

Answer 1

The correct option is A $10$

Let the conductor reaches at $y$ from the top at a time, $t$
From the forces on the conductor, $mg-BIL=ma$ ----------------(1)
Further, $E\left(EMF\right)=BvL$
Also, $E=\frac{Q}{C}$
Therefore, $Q=CBvL$

Differentiating both sides, we get

$I=CBaL$ ----------------- (2)

From (1) & (2) above
$mg=B\left(CBaL\right)L+ma$
Or, $a=\frac{mg}{m+C{B}^2{L}^2}$

$a=5m/s^2$
Since initial velocity, u=0
Therefore, $y=\frac{a{t}^2}{2}$

$y=0.5\times 5\times 4=10m$