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. $2\,R$
• D. $3\,R$

Answer 1

Answer: (B)

$R \sqrt{3}$

Answer 2

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