Question

At what distance on the axis, from the center of a circular current carrying coil of radius $r$, the magnetic field becomes $\frac{1}{8}th$ of the magnetic field at center ?

• A. $\sqrt{2}r$
• B. $2^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}r$
• C. $r\sqrt{3}$
• D. $3\sqrt{2}r$

Answer 1

The correct option is C

$r\sqrt{3}$

Step 1: Given data

Radius of circular current carrying coil $=r$.

To find the distance at which $B_{axis}=\frac{1}{8}{B}_{center}$

Step 2: Formula used

Magnetic field at a point in the axial line of a current carrying coil,

$B_{axis}=\frac{{\mu }_{0}I{r}^2}{2{\left(r^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}$

Magnetic field at the center,

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

where ${\mu }_0$ is the permittivity of free space, $I$ is current, $r$ is radius of the loop and $x$ is the distance to the point.

Step 3: Calculation

It is given that $B_{axis}=\frac{1}{8}{B}_{center}$

Therefore,

$$\begin{gathered} \frac{{\mu }_{0}I{r}^2}{2{\left(r^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{8}\frac{{\mu }_{0}I}{2r} \\ \Rightarrow \frac{r^2}{{\left(r^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{8r} \\ \Rightarrow \frac{r^3}{{\left(r^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{2^3} \\ \Rightarrow \frac{r}{\sqrt{r^2+x^2}}=\frac{1}{2} \\ \Rightarrow \frac{1}{r^2+x^2}=\frac{1}{4r^2} \\ \Rightarrow 4r^2=r^2+x^2 \\ \Rightarrow x=r\sqrt{3} \end{gathered}$$

The distance at which the magnetic field becomes $\frac{1}{8}th$ of the magnetic field at center is $r\sqrt{3}$.

Option C is correct.