Question:
An electric current is flowing through a circular coil of radius \(R\). The ratio of the magnetic field at the centre of the coil and that at a distance \(2\sqrt{2}R\) from the centre of the coil and on its axis is :
An electric current is flowing through a circular coil of radius \(R\). The ratio of the magnetic field at the centre of the coil and that at a distance \(2\sqrt{2}R\) from the centre of the coil and on its axis is :
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Use \(B_{centre} = \frac{\mu_0 I}{2R}\) and \(B_{axis} = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}\).
Updated On: Oct 31, 2026
- \(2\sqrt{2}\)
- 27
- 36
- 8
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The Correct Option is B
Solution and Explanation
At the centre: \(B_c = \frac{\mu_0 I}{2R}\).
On the axis at distance \(x\): \(B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\).
With \(x = 2\sqrt2R\): \(x^2 = 8R^2\), so \(R^2 + x^2 = 9R^2\) and \((9R^2)^{3/2} = 27R^3\).
\(B_x = \frac{\mu_0 I R^2}{2\cdot27R^3} = \frac{\mu_0 I}{54R}\).
\(\frac{B_c}{B_x} = \frac{\mu_0I/2R}{\mu_0I/54R} = 27\).
Answer: (2) 27
On the axis at distance \(x\): \(B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\).
With \(x = 2\sqrt2R\): \(x^2 = 8R^2\), so \(R^2 + x^2 = 9R^2\) and \((9R^2)^{3/2} = 27R^3\).
\(B_x = \frac{\mu_0 I R^2}{2\cdot27R^3} = \frac{\mu_0 I}{54R}\).
\(\frac{B_c}{B_x} = \frac{\mu_0I/2R}{\mu_0I/54R} = 27\).
Answer: (2) 27
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