Magnetic Effects of Current and Magnetism Part 1

The magnetic field intensity at a distance r from centre on the axis of a coil of radius a carrying current i is given by $B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2\pi i a^2}{(a^2+r^2)^{3/2}}$ ...(i) The magnetic field intensity at the centre of the coil is given by $B_{\text{centre}} = \frac{\mu_0}{4\pi} \frac{2\pi i}{a}$ ...(ii) According to question, the fractional change in the magnetic fieldintensity is given by $\frac{\Delta B}{B} = 1 - \frac{B_{\text{axis}}}{B_{\text{centre}}}$ ...(iii) Substituting the values of $B_{\text{axis}}$ and $B_{\text{centre}}$ from Eqs. (i) and (ii) respectively in Eq. (iii), we get $\frac{\Delta B}{B} = 1 - \frac{ \frac{\mu_0}{4\pi} \frac{2\pi i a^2}{(a^2+r^2)^{3/2}} }{ \frac{\mu_0}{4\pi} \frac{2\pi i}{a} }$ $= 1 - \frac{a^3}{(a^2+r^2)^{3/2}}$ $= 1 - \frac{a^3}{a^3 \left(1 + \frac{r^2}{a^2}\right)^{3/2}}$ $= 1 - \left(1 + \frac{r^2}{a^2}\right)^{-3/2}$ Given that, r < a, applying Binomial expansion, $(1-x)^{-n} = 1 - x + \frac{2 x^2}{2!} - \frac{3 \times 2 x^3}{3!} \ldots (-1)^n x^n,$ $= (1 + n x)$ we get $\frac{\Delta B}{B} = 1 - \left(1 - \frac{3}{2} \frac{r^2}{a^2}\right)$ $\frac{\Delta B}{B} = 1 - 1 + \frac{3}{2} \frac{r^2}{a^2}$ $\frac{\Delta B}{B} = \frac{3}{2} \frac{r^2}{a^2}$, which is the required expression.