ExamBro
ExamBro
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 : \(\left.\sqrt{2}=1.4\right)\)

  1. A \(65\)
  2. B \(68\)
  3. C \(66\)
  4. D \(67\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(68\)

Step-by-step Solution

Detailed explanation

Magnetic fields due to both wires will be perpendicular to each other. \(B _1=\frac{\mu_0 i _1}{2 \pi d } \quad B _2=\frac{\mu_0 i _2}{2 \pi d }\) \(B _{\text {net }}=\sqrt{ B _1^2+ B _2^2} \Rightarrow \frac{\mu_0}{2 \pi d } \sqrt{ i _1^2+ i _2^2}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app