Two long parallel wires carrying currents \(8\,A\) and \(15\,A\) in opposite directions are placed at a distance of \(7\,cm\) from each other. A point \(P\) is at equidistant from both the wires such that the lines joining the point \(P\) to the wires are perpendicular to each other. The magnitude of magnetic field at \(P\) is \(............\times 10^{-6}\,T\). (Given : \(\left.\sqrt{2}=1.4\right)\)

Pick an option to check your answer.

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Answer & Solution

Correct Answer

(B) \(68\)

Detailed explanation

Magnetic fields due to both wires will be perpendicular to each other.

\(B _1=\frac{\mu_0 i _1}{2 \pi d } \quad B _2=\frac{\mu_0 i _2}{2 \pi d }\)

\(B _{\text {net }}=\sqrt{ B _1^2+ B _2^2} \Rightarrow \frac{\mu_0}{2 \pi d } \sqrt{ i _1^2+ i _2^2}\)

\(\Rightarrow \frac{4 \pi \times 10^{-7}}{2 \pi \times(7 / \sqrt{2}) \times 10^{-2}} \times \sqrt{8^2+15^2}\left( d =\frac{7}{\sqrt{2}}\,cm \right)\)

\(\Rightarrow 68 \times 10^{-6}\,T\)

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