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)

2.5 m/s²

1.4 m/s²

9.8 m/s²

0

Solution

1.4 m/s²

Due to motion of rod, emf induced across capacitor,

e = B l v

∴ Charge stored in capacitor,

Q = C ( Blv )

Current

I = $\frac{dQ}{dt}$

= C ( Blv) $\frac{dv}{dt}$

I = CBl α

Force opposing the downward motion,

F_m = BI l

∴ F_m = B (CBla) l

F_m = B² l² Ca

Net force on rod

F_net = W $-$ F_m

= mg $-$ B² l²Ca

∴ m $α$ = mg $-$ B² l²C $α$

⇒ $α = \frac{m g}{(m + B^{2} l^{2} C)}$

so $α = \frac{0.1 \times 9.8}{(0.1 + 2^{2} \times 1^{2} \times 0.15)}$

$α$ = 1.4 m/s²

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