# The fractional change in the magnetic field intensity at a distance

Question:

The fractional change in the magnetic field intensity at a distance ' $r$ ' from centre on the axis of current carrying coil of radius ' $a$ ' to the magnetic field intensity at the centre of the same coil is:

(Take $r 1.$\frac{3}{2} \frac{\mathrm{a}^{2}}{\mathrm{r}^{2}}$2.$\frac{2}{3} \frac{a^{2}}{r^{2}}$3.$\frac{2}{3} \frac{r^{2}}{a^{2}}$4.$\frac{3}{2} \frac{r^{2}}{a^{2}}$Correct Option: , 4 Solution:$\mathrm{B}_{\text {axis }}=\frac{\mu_{0} 1 R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}B_{\text {centre }}=\frac{\mu_{0} i}{2 R}\therefore B_{\text {centre }}=\frac{\mu_{0} i}{2 a}\therefore \mathrm{B}_{\mathrm{axis}}=\frac{\mu_{0} \mathrm{ia}^{2}}{2\left(\mathrm{a}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}\therefore$fractional change in magnetic field$=\frac{\frac{\mu_{0} \mathrm{i}}{2 \mathrm{a}}-\frac{\mu_{0} \mathrm{ia}^{2}}{2\left(\mathrm{a}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}}{\frac{\mu_{0} \mathrm{i}}{2 \mathrm{a}}}=1-\frac{1}{\left[1+\left(\frac{\mathrm{r}^{2}}{\mathrm{a}^{2}}\right)\right]^{3 / 2}}\approx 1-\left[1-\frac{3}{2} \frac{r^{2}}{a^{2}}\right]=\frac{3}{2} \frac{r^{2}}{a^{2}}$Note :$\left(1+\frac{\mathrm{r}^{2}}{\mathrm{a}^{2}}\right)^{-3 / 2} \approx\left(1-\frac{3}{2} \frac{\mathrm{r}^{2}}{\mathrm{a}^{2}}\right)$[True only if$\mathrm{r} \ll<\mathrm{a}\$ ]

Hence option (4) is the most suitable option