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(B) The force per unit length between two parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Given

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$5 	imes 10^{-6}$,attractive

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(B) The force per unit length between two parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Given: $I_1 = I_2 = 5 	ext{ A}$,$d = 1 	ext{ m}$,and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.Substituting the values: $\frac{F}{l} = \frac{4\pi 	imes 10^{-7} 	imes 5 	imes 5}{2 \pi 	imes 1}$.Simplifying the expression: $\frac{F}{l} = 2 	imes 10^{-7} 	imes 25 = 50 	imes 10^{-7} 	ext{ N/m}$.Thus,$\frac{F}{l} = 5 	imes 10^{-6} 	ext{ N/m}$.Since the currents are in the same direction,the force between the wires is attractive.

Two parallel very long straight wires carrying current of $5 \text{ A}$ each are kept at a separation of $1 \text{ m}$. If the currents are in the same direction,the force per unit length between them is . . . . . . $\text{N/m}$. $(\mu_0 = 4\pi \times 10^{-7} \text{ SI units})$

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