The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis

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

a (a) Field at the centre ${B_1} = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2\pi in}}{r} = \frac{{{\mu _0}}}{2}.\frac{{ni}}{r}$ Field at a distance $h$ from the centre ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni{r^2}}}{{{{({r^2} + {h^2})}^{3/2}}}} = \frac{{{\mu _0}}}{2}.\frac{{ni{r^2}}}{{{r^3}{{\left( {1 + \frac{{{h^2}}}{{{r^2}}}} \right)}^{3/2}}}}$ $ = {B_1}{\left( {1 + \frac{{{h^2}}}{{{r^2}}}} \right)^{ - 3/2}} = {B_1}\left( {1 - \frac{3}{2}.\frac{{{h^2}}}{{{r^2}}}} \right)$(By binomial theorem) Hence $B_2$ is less than $B_1$ by a fraction $ = \frac{3}{2}\frac{{{h^2}}}{{{r^2}}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$\frac{3}{2}\frac{{{h^2}}}{{{r^2}}}$
B$\frac{2}{3}\frac{{{h^2}}}{{{r^2}}}$
C$\frac{3}{2}\frac{{{r^2}}}{{{h^2}}}$
D$\frac{2}{3}\frac{{{r^2}}}{{{h^2}}}$

Diffcult

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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