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(C) For the second wire to float in air,the downward gravitational force must be balanced by the upward magnetic force exerted by the first wire.$1$.

Vedclass · Accepted Answer

$49\, A$ (direction of current is same to first wire)

Vedclass · Answer

(C) For the second wire to float in air,the downward gravitational force must be balanced by the upward magnetic force exerted by the first wire.$1$. The gravitational force per unit length is $F_g/l = (m/l)g = 10^{-2} 	imes 9.8 = 0.098\, N/m$.$2$. The magnetic force per unit length between two parallel wires is given by $F_m/l = \frac{\mu_0}{4\pi} \cdot \frac{2 i_1 i_2}{a}$.$3$. Equating the two forces: $(m/l)g = \frac{\mu_0}{4\pi} \cdot \frac{2 i_1 i_2}{a}$.$4$. Substituting the values: $10^{-2} 	imes 9.8 = 10^{-7} 	imes \frac{2 	imes 200 	imes i}{2 	imes 10^{-2}}$.$5$. Solving for $i$: $0.098 = 10^{-7} 	imes \frac{400 i}{0.02} = 10^{-7} 	imes 20000 i = 2 	imes 10^{-3} i$.$6$. $i = \frac{0.098}{2 	imes 10^{-3}} = 49\, A$.$7$. Since the magnetic force must be upward (repulsive) to counteract gravity,the currents in the two parallel wires must flow in the same direction.

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