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Question

(A) The magnetic field at the centre of the coil is given by $B_1 = \frac{\mu_0 ni}{2r}$.The magnetic field at a distance $h$ on the axis of the coil

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$\frac{3}{2}\frac{h^2}{r^2}$

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(A) The magnetic field at the centre of the coil is given by $B_1 = \frac{\mu_0 ni}{2r}$.The magnetic field at a distance $h$ on the axis of the coil is given by $B_2 = \frac{\mu_0 ni r^2}{2(r^2 + h^2)^{3/2}}$.We can rewrite $B_2$ as $B_2 = \frac{\mu_0 ni r^2}{2r^3(1 + h^2/r^2)^{3/2}} = B_1(1 + h^2/r^2)^{-3/2}$.Using the binomial expansion $(1 + x)^n \approx 1 + nx$ for small $x$,where $x = h^2/r^2$ and $n = -3/2$,we get $B_2 \approx B_1(1 - \frac{3}{2}\frac{h^2}{r^2})$.The fractional decrease is given by $\frac{B_1 - B_2}{B_1} = 1 - \frac{B_2}{B_1} = 1 - (1 - \frac{3}{2}\frac{h^2}{r^2}) = \frac{3}{2}\frac{h^2}{r^2}$.

The magnetic field normal to the plane of a wire coil of $n$ turns and radius $r$ carrying a current $i$ is measured on the axis of the coil at a small distance $h$ from the centre. By what fraction is this field smaller than the field at the centre?

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