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A long straight wire carries a current $I_{0}$ . At distances a and b from it there are two other wires, parallel to the former one, which are interconnected by a resistance R. A conector slides without friction along the wires with a constant velocity V. Assuming the resistances of the wires, the connector, the sliding contacts, and the self-inductance of the frame to be negligible, find the force required to maintain the connector's velocity constant.

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F = {(\frac{μ_{0} I_{0}}{2 π} \ln b/a)}^{2}

F = \frac{V}{R} (\frac{μ_{0} I_{0}}{2 π} \ln b/a)

F = \frac{V}{R} {(\frac{μ_{0} I_{0}}{2 π})}^{2}

F = \frac{V}{R} {(\frac{μ_{0} I_{0}}{2 π} \ln b/a)}^{2}

answer is D.

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Detailed Solution

$d ε = B ϑ d x$

$\int d ε = \int_{a}^{b} \frac{μ_{0} i_{0}}{2 π x} ϑ d x$

$ε = \frac{μ_{0} i_{0}}{2 π} ϑ {[l n x]}_{a}^{b}$

$ε = \frac{μ_{0} i_{0}}{2 π} ϑ l n (b/a)$

$Power P = F ϑ$

$F = \frac{P}{ϑ} = \frac{ε^{2}}{R ϑ}$

$F = {[\frac{μ_{0}}{2 π} i_{0} l n (b/a)]}^{2} \frac{ϑ}{R}$

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