ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

Two identical circular wires of radius \(20\,cm\) and carrying current \(\sqrt{2}\,A\) are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wire is \(.............\times 10^{-8}\,T\). (Take \(\pi=3.14\) )

  1. A \(689\)
  2. B \(546\)
  3. C \(487\)
  4. D \(628\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(628\)

Step-by-step Solution

Detailed explanation

\(\text { Magnetic field } B_C \text { at center }=\frac{\mu_0 i }{2 r }\) \(=\frac{4 \pi \times 10^{-7}}{2 \times 0.2} \times \sqrt{2}\,T\) Net magnetic field is…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app