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(D) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $d$ is given by $F = \frac{\mu_0 i_1

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(D) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $d$ is given by $F = \frac{\mu_0 i_1 i_2 l}{2 \pi d}$.Given that $i_1 = i_2 = i$,the formula becomes $F = \frac{\mu_0 i^2 l}{2 \pi d} = 2 	imes 10^{-7} 	imes \frac{i^2 l}{d}$.Substituting the given values: $30 	imes 10^{-7} = 2 	imes 10^{-7} 	imes \frac{i^2 	imes 9}{0.15}$.Simplifying the equation: $30 = 2 	imes \frac{i^2 	imes 9}{0.15}$.$30 = \frac{18 i^2}{0.15}$.$i^2 = \frac{30 	imes 0.15}{18} = \frac{4.5}{18} = 0.25$.Therefore,$i = \sqrt{0.25} = 0.5 \, A$.

Two parallel wires of length $9 \, m$ each are separated by a distance $0.15 \, m$. If they carry equal currents in the same direction and exert a total force of $30 \times 10^{-7} \, N$ on each other,then the value of current must be ........ $A$. (in $.5$)

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