As shown in figure, two vertical conducting rails separated by distance $1.0 \mathrm{~m}$ are placed…
As shown in figure, two vertical conducting rails separated by distance $1.0 \mathrm{~m}$ are placed parallel to $z$-axis. At $z=0$, a capacitor of $0.15 \mathrm{~F}$ is connected between the rails and a metal rod of mass $100 \mathrm{gm}$ placed across the rails slides down along the rails. If a constant magnetic fields of $2.0 \mathrm{~T}$ exists perpendicular to the plane of the rails, what is the acceleration of the rod?
(Take $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )
$2.5 \mathrm{~m} / \mathrm{s}^2$
$1.4 \mathrm{~m} / \mathrm{s}^2$
$9.8 \mathrm{~m} / \mathrm{s}^2$
$0$
Solution
Due to motion of rod, emf induced across capacitor, $\varepsilon=B l v$
$\therefore \quad$ Charge stored in capacitor, $Q=C(B l v)$
$I=\frac{d Q}{d t}=C B l \frac{d v}{d t}=C B l a$
Force opposing the downward motion, $F_m=B I l$
$\therefore \quad F_m=B(C B l a) l=B^2 l^2 C a$
Net force on rod, $F_{\text {net }}=W-F_m=m g-B^2 l^2 C a$
$\therefore \quad m a=m g-B^2 l^2 C a$
or $\quad a=\frac{m g}{\left(m+B^2 l^2 C\right)}$
So, $\quad a=\frac{0.1 \times 9.8}{\left(0.1+2^2 \times 1^2 \times 0.15\right)}=1.4 \mathrm{~m} / \mathrm{s}^2$
