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Question

(A) The magnetic field on the axis of a circular coil of $N$ turns,radius $r$,and current $i$ at a distance $h$ from the center is given by: $B_{	ext

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The magnetic field on the axis of a circular coil of $N$ turns,radius $r$,and current $i$ at a distance $h$ from the center is given by: $B_{	ext{axis}} = \frac{\mu_0 N i r^2}{2(r^2 + h^2)^{3/2}}$.This can be rewritten as: $B_{	ext{axis}} = \frac{\mu_0 N i r^2}{2 r^3 (1 + h^2/r^2)^{3/2}} = \frac{\mu_0 N i}{2 r} (1 + h^2/r^2)^{-3/2}$.Using the binomial approximation $(1+x)^n \approx 1+nx$ for $h \ll r$,we get: $B_{	ext{axis}} \approx \frac{\mu_0 N i}{2 r} (1 - \frac{3h^2}{2r^2})$.The magnetic field at the center is $B_{	ext{center}} = \frac{\mu_0 N i}{2 r}$.Thus,$B_{	ext{axis}} = B_{	ext{center}} (1 - \frac{3h^2}{2r^2}) = B_{	ext{center}} - \frac{3h^2}{2r^2} B_{	ext{center}}$.The decrease in the field is $\frac{3h^2}{2r^2} B_{	ext{center}}$,so the fraction by which it is smaller is $\frac{3h^2}{2r^2}$.

The magnetic field normal to the plane of a coil of $N$ turns and radius $r$ which carries a current $i$ is measured on the axis of the coil at a distance $h$ from the centre of the coil. This is smaller than the field at the centre by the fraction,

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