Two parallel,long wires are kept 0.20 , m apa | vedclass

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(C) The formula for the force per unit length between two parallel current-carrying wires is given by:$F/L = \frac{\mu_{0} i_{1} i_{2}}{2 \pi d}$Given

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(C) The formula for the force per unit length between two parallel current-carrying wires is given by:$F/L = \frac{\mu_{0} i_{1} i_{2}}{2 \pi d}$Given:$F/L = 2 	imes 10^{-6} \, N/m$$d = 0.20 \, m$$i_{1} = i_{2} = x \, A$$\mu_{0} = 4 \pi 	imes 10^{-7} \, T \cdot m/A$Substituting the values into the formula:$2 	imes 10^{-6} = \frac{4 \pi 	imes 10^{-7} 	imes x^2}{2 \pi 	imes 0.2}$Simplifying the equation:$2 	imes 10^{-6} = \frac{2 	imes 10^{-7} 	imes x^2}{0.2}$$2 	imes 10^{-6} = 10^{-6} 	imes x^2$$x^2 = 2$$x = \sqrt{2} \approx 1.414 \, A$Thus,the value of $x$ is approximately $1.4 \, A$.

Two parallel,long wires are kept $0.20 \, m$ apart in a vacuum,each carrying a current of $x \, A$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} \, N$,then the value of $x$ is approximately:

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