Question

As shown in figure, two vertical conducting rails separated by distance $1.0m$ are placed parallel to z-axis. At $z=0$, a capacitor of $0.15F$ is connected between the rails and a metal rod of mass $100gm$ placed across the rails slides down along the rails. If a constant magnetic fields of $2.0T$ exists perpendicular to the plane of the rails, what is the acceleration of the rod?
(Take $g=9.8m/s^2)$.

• A. $2.5m/s^2$
• B. $1.4m/s^2$
• C. $9.8m/s^2$
• D. $0$

Answer 1

Answer: (B)

$1.4m/s^2$

Due to motion of rod, emf induced across capacitor, $ϵ=Blv$
$\therefore$ Charge stored in capacitor, $Q=C\left(Blv\right)$
$I=\frac{dQ}{dt}=CBl\frac{dv}{dt}=CBla$
Force opposing the downward motion, $F_m=BIl$
$\therefore F_m=B\left(CBla\right)l=B^2{l}^{2}Ca$
Net force on rod, $F_{net}=W-F_m=mg-B^2{l}^{2}CA$
$\therefore ma=mg-B^2{l}^{2}Ca$
or $a=\frac{mg}{(m+B^2{l}^{2}C)}$
So, $a=\frac{0.1\times 9.8}{(0.1+2^2\times 1^2\times 0.15)}=1.4m/s^2$.