Explanation

(b) From Biot-Savart’s law the magnetic field $(B)$ due to a conductor carrying current $I,$ at a distance $r_{1}$ is
$B_{1} = \frac{\mu_{0}I_{1}}{2\pi r_{1}}$
Magnetic field at $P$ due to current in second conductor is
$B_{2} = \frac{\mu_{0}I_{2}}{2\pi\left( r_{1} + d \right)}$
From Fleming’s right hands rule the fields at $P$ are directed opposite.
$\therefore Resultants,\ field\ B_{1} = B_{2}$
$\therefore\ \ \ \ \frac{\mu_{0}I_{1}}{2\pi r_{1}} = \ \frac{\mu_{0}I_{2}}{2\pi\left( r_{1} + d \right)}$
Given, $I_{1} = 10\ A,\ r_{1} = 5,\ r_{1} + d = 5 + 10 = 15\ cm$
$\therefore\ \ \ \ \ \ \ \ \ I_{2} = \frac{I_{1}}{r_{1}}\ \times \left( r_{1} + d \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ I_{2} = \frac{10}{5} \times 15 = 30\ A$
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