Question

The radius of a circular current-carrying coil is $R$. At what distance from the center of the coil on its axis, the intensity of the magnetic field will be $\frac{1}{2\sqrt{2}}$ times at the center?

Answer 1

Step 1: Given data

The magnetic field at the point of the axis = $\frac{1}{2\sqrt{2}}$ times the magnetic field at the center of the coil.

Step 2: Formula used

The magnetic field at the center of the coil on its axis,

$B_o=\frac{{\mu }_0}{2}\frac{I}{R}$

(where ${\mu }_0$is the magnetic permeability of free space, $I$ is the current through the coil, and $R$ is the radius of the coil)

The magnetic field at a point on the axis of the coil at a distance $x$ from the center of the coil,

$B_p=\frac{{\mu }_0}{2}\frac{I\times R^2}{{\left(R^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}$

Step 3: Calculation

$$\begin{gathered} B_p=\frac{1}{2\sqrt{2}}{B}_o \\ \frac{{\mu }_0}{2}\frac{I\times R^2}{{\left(R^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{2\sqrt{2}}\frac{{\mu }_0}{2}\frac{I}{R} \end{gathered}$$

Equating both sides, we get

$2\sqrt{2}{R}^3={\left(R^2+x^2\right)}^{\frac{3}{2}}$

Squaring both sides, we get

$8R^6={\left(R^2+x^2\right)}^3$

Taking cube root both sides, we get

$$\begin{gathered} 2R^2=R^2+x^2 \\ {R}^2=x^2 \\ R=x \end{gathered}$$

At distance ‘R’ from the center of the coil on its axis, the intensity of the magnetic field will be $\frac{1}{2\sqrt{2}}$times at the center.