Question

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

• A. $\frac{2}{3}\frac{h^2}{r^2}$
• B. $\frac{3}{2}\frac{r^2}{h^2}$
• C. $\frac{3}{2}\frac{h^2}{r^2}$
• D. $\frac{1}{3}\frac{h^2}{r^2}$

Answer 1

Answer: (C)

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

Magnetic field at the centre $B_c=\frac{{\mu }_{0}ni}{2r}$
Magnetic field on the axis at a small distance h from the centre,

$B_h=\frac{2n{\mu }_{0}i\cdot \pi r^2}{4\pi (r^2+h^2{)}^{\frac{3}{2}}}$
$=\frac{n{\mu }_{0}i\cdot r^2}{2(r^2+h^2{)}^{\frac{3}{2}}}$
$=\left(\frac{n{\mu }_{0}i}{2r}\right){(1+\frac{h^2}{r^2})}^{\frac{-3}{2}}$
$\approx \left(\frac{n{\mu }_{0}i}{2r}\right)(1+\left(\frac{-3}{2}\right)\left(\frac{h^2}{r^2}\right))$

$\therefore$ $\frac{B_c-B_h}{B_c}=\frac{3h^2}{2r^2}$