91Ó°ÊÓ

The magnetic field at the axis of the circular coil is given by

$B_x=\frac{{\mathrm{μ}}_{0}NI{R}^2}{2\left(x^2+R^2\right)}$

The magnetic field at center of loop

The magnetic field at center of loop is given by

$B_c=\frac{{\mathrm{μ}}_{0}I}{2R}âˆ\pounds$

Where, B is the magnetic field due to wire, ${\mathrm{μ}}_0$is the permeability of vaccum, l is the current through the wire and R is the distance from wire.

Given : The radius of circular coil is R = 2.40 cm .

The total number of turns in coil is N = 800.

Magnetic field at a point x from coil$\left(B_x\right)=$half of the magnetic field at center of coil$\left(B_c\right)$ .

$\begin{array}{c}{B}_x=\frac{1}{2}\left(B_c\right)\\ \frac{{\mathrm{μ}}_{0}NI{R}^2}{2\left(x^2+R^2\right)}=\frac{1}{2}×\frac{{\mathrm{μ}}_{0}I}{2R}\\ 2N{R}^3=x^2+R^2\end{array}$

$x=\sqrt{2N{R}^3−R^2}$

Now, putting the values of constants in above equation

$\begin{array}{r}x=\sqrt{2×800×(2.40\mathrm{cm}{)}^3−(2.40\mathrm{cm}{)}^2}\\ x=1.84\mathrm{cm}\end{array}$

Thus,the distance on the axis of coil at which magnetic field is half of the value of magnetic field at center of the coil is x = 1.84 cm .