Question

A circular coil carrying current 'I' has radius 'R' and magnetic field at the centre is 'B' . At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?

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

Answer 1

Answer: (B)

According to the question, magnetic field at the centre of circular coil is given by
$B_{\text{centre }}=\frac{{\mu }_{0}Ni}{2R}...(i)$
$B_{\text{axis }}=\frac{{\mu }_0}{4\pi }\frac{2\pi Ni{R}^2}{{\left(x^2+R^2\right)}^{3/2}}$
$\frac{B}{8}=\frac{{\mu }_{0}Ni{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}$
From Eq. (i), we get
$\frac{{\mu }_{0}Ni}{8\times 2R}=\frac{{\mu }_{0}Ni{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}$
$\frac{1}{8R}=\frac{1}{{\left(x^2+R^2\right)}^{3/2}}$
$\frac{1}{8R^3}=\frac{1}{{\left(x^2+R^2\right)}^{3/2}}$
$8R^3={\left(x^2+R^2\right)}^{3/2}$ (on taking cube root)
$2R={\left(x^2+R^2\right)}^{1/2}$ (on taking square root)
$4R^2=\left(x^2+R^2\right)$
$4R^2-R^2=x^2$
$3R^2=x^2$
$x=R\sqrt{3}$