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(B) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_{	ext{centre}} = \frac{\mu_0 I}{2R}$.The ma

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$27$

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(B) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_{	ext{centre}} = \frac{\mu_0 I}{2R}$.The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_{	ext{axis}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.The ratio of the magnetic field at the centre to that at the axis is:$\frac{B_{	ext{centre}}}{B_{	ext{axis}}} = \frac{\mu_0 I / 2R}{\mu_0 I R^2 / 2(R^2 + x^2)^{3/2}} = \frac{(R^2 + x^2)^{3/2}}{R^3} = \left(1 + \frac{x^2}{R^2}ight)^{3/2}$.Given $x = 2\sqrt{2}R$,we substitute this into the ratio:$\frac{B_{	ext{centre}}}{B_{	ext{axis}}} = \left(1 + \frac{(2\sqrt{2}R)^2}{R^2}ight)^{3/2} = \left(1 + \frac{8R^2}{R^2}ight)^{3/2} = (1 + 8)^{3/2} = (9)^{3/2} = (3^2)^{3/2} = 3^3 = 27$.

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 on its axis is:

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