The ratio of the magnetic field at the centre | vedclass

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

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$2\sqrt{2}$

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(D) The magnetic field at the centre of a circular coil of radius $a$ carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2a}$.The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_a = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$.Given $x = a$,the magnetic field at distance $a$ is $B_a = \frac{\mu_0 I a^2}{2(a^2 + a^2)^{3/2}} = \frac{\mu_0 I a^2}{2(2a^2)^{3/2}} = \frac{\mu_0 I a^2}{2(2\sqrt{2} a^3)} = \frac{\mu_0 I}{4\sqrt{2} a}$.The ratio of the magnetic field at the centre to the field at distance $a$ is $\frac{B_c}{B_a} = \frac{\mu_0 I / 2a}{\mu_0 I / 4\sqrt{2} a} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

The ratio of the magnetic field at the centre of a current-carrying coil of radius $a$ to the magnetic field at a distance $a$ from the centre of the coil along its axis is

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