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)

Question

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) (A) 2.5 m/s2 (B) 1.4 m/s2 (C) 9.8 m/s2 (D) 0

Tardigrade Admin · Accepted Answer

Due to motion of rod, emf induced across capacitor,

ε

= Blv

∴

Charge stored in capacitor, Q = C (Blv)

I = \frac{d Q}{d t} = CBl \frac{d v}{d t} = CBl a

Force opposing the downward motion,

F_{m}

= BIl

∴ F_{m} = B (CBl a) l = B^{2} l^{2}

Ca
Net force on rod,

F_{n e t} = W - F_{m} = m g - B^{2} l^{2}

Ca
or

a = \frac{m g}{( m + B ^{2} l ^{2} C a )}

So,

a = \frac{0.1 \times 9.8}{( 0.1 + 2 ^{2} \times 1 ^{2} \times 0.15 )} = 1.4 m / s^{2}

2.5 m/s²

1.4 m/s²

9.8 m/s²

0