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

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.

  1. A \(9: 4\)
  2. B \(12: \sqrt{5}\)
  3. C \(8: 1\)
  4. D \(5: \sqrt{3}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(8: 1\)

Step-by-step Solution

Detailed explanation

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