The magnitude of the magnetic induction (B) at a point on the axis...

The magnetic field on the axis of a circular coil of radius $R$ at a distance $r$ from its centre is given by:

$B = \dfrac{\mu_0 n i R^2}{2(R^2 + r^2)^{3/2}}$

For a large distance $r \gg R$, the expression simplifies to:

$B = \dfrac{\mu_0 n i R^2}{2r^3}$

Multiplying the numerator and denominator by $\pi$, we get:

$B = \dfrac{\mu_0 n i (\pi R^2)}{2\pi r^3}$

Since the area of the coil is $A = \pi R^2$, we can write:

$B = \dfrac{\mu_0 n i A}{2\pi r^3}$

Substituting $\mu_0 = 4\pi \times 10^{-7}$ SI units:

$B = \dfrac{4\pi \times 10^{-7} \times n i A}{2\pi r^3} = \dfrac{2 n i A}{r^3} \times 10^{-7}$

Comparing this with $B = x \times 10^{-7}$, we find:

$x = \dfrac{2 n i A}{r^3}$

Answer: $\dfrac{2niA}{r^3}$