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(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by $F/l = \fr

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by $F/l = \frac{\mu_0 I_1 I_2}{2 \pi h}$.Since the currents are in opposite directions,the force is repulsive.At equilibrium,the repulsive magnetic force balances the gravitational force acting on wire $PQ$:$F = mg$$\frac{\mu_0 I^2 l}{2 \pi h} = mg$Given: $I = 25 \,A$,$l = 1 \,m$,$m = 2.5 \,g = 2.5 	imes 10^{-3} \,kg$,$g = 9.8 \,m/s^2$,$\mu_0 = 4 \pi 	imes 10^{-7} \,T \cdot m/A$.Substituting the values:$h = \frac{\mu_0 I^2 l}{2 \pi m g} = \frac{(4 \pi 	imes 10^{-7}) 	imes (25)^2 	imes 1}{2 \pi 	imes (2.5 	imes 10^{-3}) 	imes 9.8}$$h = \frac{2 	imes 10^{-7} 	imes 625}{2.5 	imes 10^{-3} 	imes 9.8} = \frac{1250 	imes 10^{-7}}{24.5 	imes 10^{-3}} = \frac{1250}{24.5} 	imes 10^{-4} \,m \approx 51 	imes 10^{-4} \,m \approx 5.1 \,mm$.Rounding to the nearest given option,the height is $5 \,mm$.

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