Question:
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 magnetic field at P is ___ \(\times 10^{-6}\) T. (Given: \(\sqrt{2} = 1.4\), \(μ_0 = 4π\times 10^{-7}\) SI unit)
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 magnetic field at P is ___ \(\times 10^{-6}\) T. (Given: \(\sqrt{2} = 1.4\), \(μ_0 = 4π\times 10^{-7}\) SI unit)
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The point is $5$ cm from each wire, and the two fields are perpendicular.
Updated On: Oct 1, 2026
- \(62\)
- \(65\)
- \(68\)
- \(70\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the distances
Lines from P to the wires are perpendicular and equal, so the wires and P form a right isosceles triangle with hypotenuse \(7\) cm. Each distance is \(r=\frac{7}{\sqrt2}=\frac{7}{1.4}=5\) cm.
Step 2: Fields from each wire
\(B_1=\frac{\mu_0I_1}{2\pi r}=\frac{2\times10^{-7}\times8}{0.05}=3.2\times10^{-5}\) T.
\(B_2=\frac{2\times10^{-7}\times15}{0.05}=6\times10^{-5}\) T.
Step 3: Combine
Each field is perpendicular to its line from the wire. These lines are perpendicular, so \(B_1\perp B_2\).
\(B=\sqrt{(3.2)^2+(6)^2}\times10^{-5}=\sqrt{46.24}\times10^{-5}=6.8\times10^{-5}\) T \(=68\times10^{-6}\) T. Option (C).
Final Answer:
The field is \(68\times10^{-6}\) T, option (C). \[ \boxed{\text{(C) }68} \]
Lines from P to the wires are perpendicular and equal, so the wires and P form a right isosceles triangle with hypotenuse \(7\) cm. Each distance is \(r=\frac{7}{\sqrt2}=\frac{7}{1.4}=5\) cm.
Step 2: Fields from each wire
\(B_1=\frac{\mu_0I_1}{2\pi r}=\frac{2\times10^{-7}\times8}{0.05}=3.2\times10^{-5}\) T.
\(B_2=\frac{2\times10^{-7}\times15}{0.05}=6\times10^{-5}\) T.
Step 3: Combine
Each field is perpendicular to its line from the wire. These lines are perpendicular, so \(B_1\perp B_2\).
\(B=\sqrt{(3.2)^2+(6)^2}\times10^{-5}=\sqrt{46.24}\times10^{-5}=6.8\times10^{-5}\) T \(=68\times10^{-6}\) T. Option (C).
Final Answer:
The field is \(68\times10^{-6}\) T, option (C). \[ \boxed{\text{(C) }68} \]
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