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  4. As shown in figure, two vertical conducting rails separated by distance 1.0 m are placed parallel to z-axis. At z = 0, a capacitor of 0.15 F is connected between the rails and a metal rod of mass 100 gm placed across the rails slides down along the rails. If a constant magnetic fields of 2.0 T exists perpendicular to the plane of the rails, what is the acceleration of the rod? (Take g = 9.8 m/s2)

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

AIIMSAIIMS 2017Electromagnetic Induction

Solution:

Due to motion of rod, emf induced across capacitor, $\varepsilon$ = Blv
$\therefore \quad$ Charge stored in capacitor, Q = C (Blv)
$I=\frac{dQ}{dt}=CBl \frac{dv}{dt}=CBla$
Force opposing the downward motion, $F_{m}$ = BIl
$\therefore \quad F_{m} = B \left(CBla\right)l = B^{2} l^{2}$ Ca
Net force on rod, $F_{net} = W - F_{m} = mg - B^{2}l^{2}$ Ca
or$\quad a=\frac{mg}{\left(m+B^{2}l^{2}Ca\right)}$
So,$\quad a=\frac{0.1\times9.8}{\left(0.1+2^{2}\times1^{2}\times0.15\right)}=1.4\,m/s^{2}$