Question

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

• A. $2R$
• B. $\frac{3R}{2}$
• C. $R$
• D. $\frac{R}{2}$

Answer 1

The correct option is C $R$

$B_{\text{axis}}=\frac{1}{2\sqrt{2}}{B}_{\text{centre}}$

$\Rightarrow \frac{{\mu }_{0}ni{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{{\mu }_{0}ni}{2R}\times \frac{1}{2\sqrt{2}}$

$\Rightarrow \frac{R^2}{2(R^2+x^2{)}^{3/2}}=\frac{1}{4R\sqrt{2}}$

or, $2\sqrt{2}{R}^3=(R^2+x^2{)}^{3/2}$

Taking power $\frac{2}{3}$ on both sides,

$(2\sqrt{2}{)}^{2/3}{R}^2=R^2+x^2$

$\Rightarrow 2R^2=R^2+x^2$

or, $2R^2-R^2=x^2$

$\therefore x=\sqrt{R^2}=R$

Hence, the correct option is $\left(c\right)$