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Two long thin parallel conductors of the shape shown in fig carry direct currents $I_{1} and I_{2}$ . The separation between the conductors is a, the width of the right-hand conductor is equal to b. With both conductors lying in one plane, find the magnetic interaction force between them reduced to a unit of their length.

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\frac{μ_{0}}{4 π} \frac{I_{1} I_{2}}{a} I n (1 + a/b)

\frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2}}{b} I n (1 + b/a)

\frac{μ_{0}}{4 π} \frac{I_{1} I_{2}}{b} I n (1 - a/b)

\frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2}}{b} I n (1 - b/a)

answer is B.

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Detailed Solution

Consider an in fine tisional strip of unit caring a current dI

$d I = (\frac{I_{2}}{b}) d x$

The field due to wire at the strip is wire and infinetisinal element wire is

$dF = \frac{μ_{0} I_{1} (d I)}{2 π x} = \frac{μ_{0} I_{1} I_{2}}{2 π b} . \frac{d x}{x}$

$F = \int dF = \frac{μ_{0} I_{1} I_{2}}{2 π b} \int_{a}^{a + b} \frac{d x}{x}$

$= \frac{μ_{0} I_{1} I_{2}}{2 π b} \log (\frac{a + b}{a})$

$= \frac{2 μ_{0} I_{1} I_{2}}{4 π b} \log (1 + \frac{b}{a})$

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