AP EAMCET · PHYSICS · Magnetic Effects of Current
The magnetic field at the centre of a current carrying circular coil of radius \(R\) is \(B_c\) and the magnetic field at a point on its axis at a distance \(R\) from its centre is \(B_a\). The value of \(\frac{B_c}{B_a}\) is
- A \(\sqrt{2}\)
- B \(\frac{1}{2 \sqrt{2}}\)
- C \(2 \sqrt{2}\)
- D \(\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(B_c = \frac{\mu_0 I}{2R}\) \(B_a = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\) At \(x=R\): \(B_a = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2\sqrt{2}R^3)} = \frac{\mu_0 I}{4\sqrt{2}R}\)…
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