= Solution
The <force> on the real <line charge> is due to the induced conductor field, represented by the <image charge>; its own singular field must be omitted. The opposite image at distance $2d$ produces at the origin a field toward the conductor,
$$
\mathbf E_{\rm image}(0)=\frac{\mu}{4\pi\varepsilon_0d}\mathbf e_x.
$$
Multiplying by the charge per unit length gives \b[the attractive force per unit length]:
$$
\boxed{\frac{\mathbf F}{L}=\frac{\mu^2}{4\pi\varepsilon_0d}\mathbf e_x.}
$$
The attraction holds for either sign of $\mu$. The geometry assumes $d>0$, so the real line lies outside the conductor.
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