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

Vedclass · Accepted Answer

$1.6 	imes 10^{-5} \ N$

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(B) The force per unit length between two parallel current-carrying wires is given by the formula: $F = \frac{\mu_0 I_1 I_2 l}{2 \pi d}$.Given:$I_1 = 10 \ A$$I_2 = 4 \ A$$d = 2 \ cm = 2 	imes 10^{-2} \ m$$l = 4 \ cm = 4 	imes 10^{-2} \ m$$\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$Substituting the values into the formula:$F = \frac{(4 \pi 	imes 10^{-7}) 	imes 10 	imes 4 	imes (4 	imes 10^{-2})}{2 \pi 	imes (2 	imes 10^{-2})}$$F = \frac{2 	imes 10^{-7} 	imes 40 	imes 4 	imes 10^{-2}}{2 	imes 10^{-2}}$$F = 2 	imes 10^{-7} 	imes 40 	imes 2$$F = 160 	imes 10^{-7} \ N = 1.6 	imes 10^{-5} \ N$.Thus,the correct option is $B$.

Two long and parallel straight wires $A$ and $B$ carrying currents of $10 \ A$ and $4 \ A$ in the same direction are separated by a distance of $2 \ cm$. Estimate the force on a $4 \ cm$ section of wire $A$. ( $\mu_0 = 4 \pi \times 10^{-7} \ SI$ ).

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