At what distance on the axis,from the centre | vedclass

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Question

(C) The magnetic field at the centre of a circular coil is given by $B_{centre} = \frac{\mu_0 I}{2r}$.The magnetic field at a distance $x$ on the axis

Vedclass · Accepted Answer

$\sqrt{3} r$

Vedclass · Answer

(C) The magnetic field at the centre of a circular coil is given by $B_{centre} = \frac{\mu_0 I}{2r}$.The magnetic field at a distance $x$ on the axis of the coil is given by $B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.According to the problem,$B_{axis} = \frac{1}{8} B_{centre}$.Substituting the expressions: $\frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 I}{2r} \right)$.Simplifying the equation: $\frac{r^2}{(r^2 + x^2)^{3/2}} = \frac{1}{8r}$.$(r^2 + x^2)^{3/2} = 8r^3$.Taking the power of $2/3$ on both sides: $r^2 + x^2 = (8r^3)^{2/3} = 4r^2$.$x^2 = 3r^2$.$x = \sqrt{3} r$.

At what distance on the axis,from the centre of a circular current-carrying coil of radius $r$,does the magnetic field become $1/8$th of the magnetic field at the centre?

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