The magnetic field at the centre of a current | vedclass

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

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$2 \sqrt{2}$

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(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2R}$.The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_a = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given $x = R$,we substitute this into the formula for $B_a$:$B_a = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2^{3/2} R^3)} = \frac{\mu_0 I}{2 \cdot 2 \sqrt{2} R} = \frac{\mu_0 I}{4 \sqrt{2} R}$.Now,we find the ratio $\frac{B_c}{B_a}$:$\frac{B_c}{B_a} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{4 \sqrt{2} R}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$.Therefore,the correct option is $C$.

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

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