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(B) The force per unit length between two parallel conductors carrying currents $i_1$ and $i_2$ separated by a distance $x$ is given by $F = \frac{\mu

Vedclass · Accepted Answer

$\frac{\mu_0}{2 \pi} i_1 i_2 \ln(2)$

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(B) The force per unit length between two parallel conductors carrying currents $i_1$ and $i_2$ separated by a distance $x$ is given by $F = \frac{\mu_0 i_1 i_2}{2 \pi x}$.Since the currents are in the same direction,the force is attractive.To increase the distance from $d$ to $2d$,we must perform work against this attractive force.The work done per unit length is given by the integral of the force with respect to distance:$W = \int_{d}^{2d} F \, dx = \int_{d}^{2d} \frac{\mu_0 i_1 i_2}{2 \pi x} \, dx$$W = \frac{\mu_0 i_1 i_2}{2 \pi} [\ln(x)]_{d}^{2d}$$W = \frac{\mu_0 i_1 i_2}{2 \pi} (\ln(2d) - \ln(d))$$W = \frac{\mu_0 i_1 i_2}{2 \pi} \ln\left(\frac{2d}{d}\right) = \frac{\mu_0 i_1 i_2}{2 \pi} \ln(2)$.

Two long straight parallel conductors are carrying currents $i_1$ and $i_2$ in the same direction. The work done per unit length,when the distance between them is doubled,is:

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