Question

A circular coil of wire of radius $r$ has $n$ turns and carries a current $i$. The magnetic induction $\left(B\right)$ at a point on the axis of the coil at a distance $\sqrt{3}r$ from its center is-

• A. $\frac{{\mu }_{0}ni}{4r}$
• B. $\frac{{\mu }_{0}ni}{8r}$
• C. $\frac{{\mu }_{0}ni}{16r}$
• D. $\frac{{\mu }_{0}ni}{32r}$

Answer 1

The correct option is C $\frac{{\mu }_{0}ni}{16r}$

The magnetic induction at an axial point,

$B=\frac{{\mu }_{0}ni{r}^2}{2{(r^2+x^2)}^{\frac{3}{2}}}[\because x=\sqrt{3}r]$

$=\frac{{\mu }_{0}ni{r}^2}{2(r^2+(\sqrt{3}r{)}^2{)}^{\frac{3}{2}}}$

$=\frac{{\mu }_{0}ni{r}^2}{2(4r^2{)}^{\frac{3}{2}}}$

$=\frac{{\mu }_{0}ni{r}^2}{2(2r{)}^3}$

$=\frac{{\mu }_{0}ni{r}^2}{16r^3}$

$\therefore B=\frac{{\mu }_{0}ni}{16r}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.