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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}}}{{4\pi }} \cdot \fr

Vedclass · Accepted Answer

Repulsive force ${10^{ - 4}} 	ext{ N/m}$

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(B) The force per unit length between two parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2{i_1}{i_2}}}{r}$ Here,${i_1} = 10 	ext{ A}$,${i_2} = 5 	ext{ A}$,and $r = 0.1 	ext{ m}$. Since the currents are in opposite directions,the force is repulsive. Substituting the values: $\frac{F}{l} = {10^{ - 7}} 	imes \frac{{2 	imes 10 	imes 5}}{{0.1}} = {10^{ - 7}} 	imes \frac{{100}}{{0.1}} = {10^{ - 7}} 	imes 1000 = {10^{ - 4}} 	ext{ N/m}$. Thus,the force is a repulsive force of ${10^{ - 4}} 	ext{ N/m}$.

What is the force per unit length between the two wires shown in the figure? [${\mu _0} = 4\pi \times {10^{ - 7}} \text{ T}\,m/A$]

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