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