The magnetic field due to a current-carrying | vedclass

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(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from the centre is given by $B_{axis} = \frac{\mu_0 I R^2}{2(R^2

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$27$

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(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from the centre is given by $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given the diameter $d = 2R$, so the radius $R = d/2$.The distance $x = \sqrt{2} d = 2\sqrt{2} R$.Substituting these into the formula:$B = \frac{\mu_0 I R^2}{2(R^2 + (2\sqrt{2} R)^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2 + 8R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(9R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(27 R^3)} = \frac{\mu_0 I}{54 R}$.The magnetic field at the centre of the coil is $B_{centre} = \frac{\mu_0 I}{2R}$.Comparing the two expressions:$B_{centre} = \frac{\mu_0 I}{2R} = 27 	imes \left( \frac{\mu_0 I}{54 R} ight) = 27 B$.

The magnetic field due to a current-carrying circular coil on its axis at a distance of $\sqrt{2} \,d$ from the centre of the coil is $B$. If $d$ is the diameter of the coil, then the magnetic field at the centre of the coil is: (in $B$)

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