Question

At very close point on the axis of a current carrying circular coil of radius 'R', the value of magnetic field decreases by a fraction of $5\%$ with respect to centre value. Find the position of the point from the centre of the coil.

Answer 1

Axial magnetic field$=\frac{{\mu }_{0}I{r}^2}{2(r^2+x^2{)}^{3/2}}$

$B_c=\frac{{\mu }_{0}I}{2r}$

$B_a=\frac{{\mu }_{0}I{r}^2}{2(r^2+x^2{)}^{3/2}}=\frac{{\mu }_{0}I}{2r}\times \frac{95}{100}$

$=\frac{r^2}{2(r^2+x^2{)}^{3/2}}=\frac{95}{100\times 2r}$

$=\frac{r^3}{(r^2+x^2{)}^{3/2}}=\frac{95}{100}$

$x<<r$

$\frac{r^3}{2^2\times \frac{3}{2}{[1+\frac{x^2}{r^2}]}^{3/2}}=\frac{95}{100}$

$=(1+\frac{3x^2}{2r^2})=\frac{100}{95}$

$=\frac{19+5+x^2}{2r^2}=20$

$=57x^2=2r^2$

$=28.5x^2=r^2$

$x=\frac{r}{\sqrt{28.5}}$

$x=\frac{r}{5.34}$

$x=0.184r$