A horizontal wire carries 160 A current. Bel | vedclass

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(A) Since the current in both wires flows in the same direction, the magnetic force between them is attractive. To keep the lower wire suspended in th

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) Since the current in both wires flows in the same direction, the magnetic force between them is attractive. To keep the lower wire suspended in the air, the upward magnetic force per unit length must balance the downward gravitational force per unit length. Let $I_1 = 160 \ A$ be the current in the upper wire, $I_2$ be the current in the lower wire, $d = 4 \ cm = 4 	imes 10^{-2} \ m$ be the distance, and $\lambda = 10 \ g \ m^{-1} = 10 	imes 10^{-3} \ kg \ m^{-1}$ be the linear mass density. The condition for equilibrium is: $F_{magnetic} = F_{gravitational}$ $\frac{\mu_0 I_1 I_2}{2 \pi d} = \lambda g$ Rearranging for $I_2$: $I_2 = \frac{2 \pi d \lambda g}{\mu_0 I_1}$ Substituting the given values: $I_2 = \frac{2 \pi 	imes (4 	imes 10^{-2}) 	imes (10 	imes 10^{-3}) 	imes 10}{4 \pi 	imes 10^{-7} 	imes 160}$ $I_2 = \frac{8 \pi 	imes 10^{-3} 	imes 10}{4 \pi 	imes 10^{-7} 	imes 160} = \frac{8 	imes 10^{-2}}{4 	imes 10^{-7} 	imes 160} = \frac{2 	imes 10^5}{160} = \frac{200000}{160} = 125 \ A$ Thus, the current in the lower wire is $125 \ A$.

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