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Find the force on the conductor carrying current `I_1` as shown in figure.
image

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Correct Answer - `(mu_0)/(4pi)2II_1log_e((1+x)/(x))`
the magnetic field induction at a point distance r from the long linear conductor carrying current is
image
`B=(mu_0)/(4pi)(2I)/(r)`
Force on the current element of length dr carrying current `I_1` is
`dF=(mu_0)/(4pi)(2I)/(r)I_1dr=(mu_0)/(4pi)2II_1(dr)/(r)`
Total force on the conductor of length l placed perpendicularly is
`F=(mu_0)/(4pi)2II_1int_(x)^((l+x))(dr)/(r)=(mu_0)/(4pi)2II_1log_e((l+x))/(x)`

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