Question

The radius of a circular current carrying coil is r. If the fractional decrease in field value with respect to the centre of the coil for a near by axial point is 1% then find axial position of that point.

Answer 1

Sipmle but you know the binomial thorem and its uses

Answer 2

Given : The radius of a circular current carrying coil is r. If the fractional decrease in field value with respect to the centre of the coil for a near by axial point is 1% .

To find : The axial position of that point.

solution : magnetic field at a point on axial line is given by, $B_{axial}=\frac{\mu_0Ir^2}{2(x^2+r^2)^{3/2}}$

here x << r so, x/r << 1

from binomial expansion,

$B_{axial}=\frac{\mu_0I}{2r}\left(1-\frac{3x^2}{2r^2}\right)$.......(1)

magnetic field at the centre of circular coil, $B_{centre}=\frac{\mu_0I}{2r}$......(2)

given, $\Delta B$ = 1% × $B_{centre}$

⇒$B_{centre}-B_{axial}=\frac{B_{centre}}{100}$

⇒$\frac{\mu_0I}{2r}-\frac{\mu_0I}{2r}\left(1-\frac{3x^2}{2r^2}\right)=\frac{1}{100}\frac{\mu_0I}{2r}$

⇒$\frac{3x^2\mu_0I}{4r^3}=\frac{1}{100}\frac{\mu_0I}{2r}$

⇒$\frac{3x^2}{2r^2}=\frac{1}{100}$

⇒x² = r²/150

⇒x = ± r/√150

Therefore axial positions of that point are r/√150 or -r/√150