Question

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

• A. $\frac{2h^2}{13r^2}$
• B. $\frac{3r^2}{2h^2}$
• C. $\frac{3h^2}{2r^2}$
• D. $\frac{2h^2}{3r^2}$

Answer 1

The correct option is C $\frac{3h^2}{2r^2}$

Magnetic field on the axis due to a current carrying coil in 3 -at a

distance in from center,

$B_1=\frac{\mu 0I{R}^2}{2(R^2+h^2{)}^{3/2}}$

and at the centre,

$B_2=\frac{\mu oI}{2R}$

so Ratio, $\frac{B_1}{B_2}=\frac{(R^2+h^2)3/2}{R^3}$

For small h, apply binomial expansion.

$\Rightarrow \frac{B_1}{B_2}=\frac{R^3+\frac{3}{2}R{h}^2+...}{R^3}\simeq 1+\frac{3}{2}.\frac{R^2}{R^2}$ (eliminate a higher order of h').

$\therefore$ Change fraction,

$\frac{\Delta B}{B_2}=\frac{B_1}{B_2-1}=\frac{3h^2}{2R^2}$ (Ans)