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Two long parallel wires carrying currents $8 \mathrm{~A}$ and $15 \mathrm{~A}$ in opposite directions are placed at a distance of $7 \mathrm{~cm}$ from each other. A point ' $\mathrm{P}$ ' is at equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of magnetic field at point ' $\mathrm{P}$ ' is $(\sqrt{2}=1.4)$ ( $\mu_0=4 \pi \times 10^{-7}$ SI units)
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Verified Answer
The correct answer is:
$68 \times 10^{-6} \mathrm{~T}$
Magnetic field produced by two wires
$\mathrm{B}_1=\frac{\mu_0 \mathrm{I}_1}{2 \pi \mathrm{X}} \text { and } \mathrm{B}_2=\frac{\mu_0 \mathrm{I}_2}{2 \pi \mathrm{X}}$
From Figure,
$\begin{aligned}
\mathrm{B}_{\text {net }} & =\sqrt{\mathrm{B}_1^2+\mathrm{B}_2^2} \\
& =\frac{\mu_0}{2 \pi \mathrm{X}} \sqrt{\mathrm{I}_1^2+\mathrm{I}_2^2}
\end{aligned}$
Also, using Pythagoras theorem, $2 \mathrm{X}^2=7 \times 7 \mathrm{~cm}$
$\begin{aligned}
\therefore \quad \mathrm{X} & =\frac{7}{\sqrt{2}} \mathrm{~cm} \\
\mathrm{~B}_{\text {net }} & =\frac{4 \pi \times 10^{-7}}{2 \pi \times \frac{7}{\sqrt{2}} \times 10^{-2}} \sqrt{15^2+8^2} \\
& \approx 68 \times 10^{-6} \mathrm{~T}
\end{aligned}$
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