JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 : \(\left.\sqrt{2}=1.4\right)\)
- A \(65\)
- B \(68\)
- C \(66\)
- D \(67\)
Answer & Solution
Correct Answer
(B) \(68\)
Step-by-step Solution
Detailed explanation
Magnetic fields due to both wires will be perpendicular to each other. \(B _1=\frac{\mu_0 i _1}{2 \pi d } \quad B _2=\frac{\mu_0 i _2}{2 \pi d }\) \(B _{\text {net }}=\sqrt{ B _1^2+ B _2^2} \Rightarrow \frac{\mu_0}{2 \pi d } \sqrt{ i _1^2+ i _2^2}\)…
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