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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2022
  • Medium
  • MCQ

The magnetic field at the center of current carrying circular loop is \(B _{1}\). The magnetic field at a distance of \(\sqrt{3}\) times radius of the given circular loop from the center on its axis is \(B_{2}\). The value of \(B_{1} / B_{2}\) will be.

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Answer & Solution

Correct Answer

(C) \(8: 1\)

Step-by-step Solution

Detailed explanation

\(B_{1}=\frac{\mu_{0} I}{2 R}\)

\(B_{2}=\frac{\mu_{0} I^{2}}{2\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{1}{8}\left(\frac{\mu_{0} I}{2 R}\right)=\frac{B_{1}}{8}\)

\(\frac{B_{1}}{B_{2}}=\frac{8}{1}\)

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