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(D) Current in wire $A$,$I_A = 8.0 \; A$.Current in wire $B$,$I_B = 5.0 \; A$.Distance between the wires,$r = 4.0 \; cm = 0.04 \; m$.Length of the sec

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$2 	imes 10^{-5} \; N$

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(D) Current in wire $A$,$I_A = 8.0 \; A$.Current in wire $B$,$I_B = 5.0 \; A$.Distance between the wires,$r = 4.0 \; cm = 0.04 \; m$.Length of the section of wire $A$,$l = 10 \; cm = 0.1 \; m$.The force per unit length between two parallel wires is given by $f = \frac{\mu_0 I_A I_B}{2 \pi r}$.The total force $F$ on a length $l$ is $F = \frac{\mu_0 I_A I_B l}{2 \pi r} = \frac{\mu_0}{4 \pi} \cdot \frac{2 I_A I_B l}{r}$.Substituting the values:$F = 10^{-7} 	imes \frac{2 	imes 8.0 	imes 5.0 	imes 0.1}{0.04}$.$F = 10^{-7} 	imes \frac{8.0}{0.04} = 10^{-7} 	imes 200 = 2 	imes 10^{-5} \; N$.Since the currents are in the same direction,the force is attractive.

Two long and parallel straight wires $A$ and $B$ carrying currents of $8.0 \; A$ and $5.0 \; A$ in the same direction are separated by a distance of $4.0 \; cm$. Estimate the force on a $10 \; cm$ section of wire $A$.

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