Question

The ratio of magnetic field at the centre of the coil of radius $R$ and at a distance $3R$ on its axis is

• A. $10\sqrt{10}$
• B. $20\sqrt{10}$
• C. $2\sqrt{10}$
• D. $\sqrt{10}$

Answer 1

The correct option is A $10\sqrt{10}$

Magnetic field at the centre due to a circular loop is,

$B_{\text{centre}}=\frac{{\mu }_{o}I}{2R}.........\left(1\right)$

Magnetic field on the axis of a circular loop is,

$B_{\text{axis}}=\frac{{\mu }_{o}I{R}^2}{2(x^2+R^2{)}^{\frac{3}{2}}}$

For, $x=3R$

$B_{\text{axis}}=\frac{{\mu }_{o}I{R}^2}{2(9R^2+R^2{)}^{\frac{3}{2}}}=\frac{{\mu }_{o}I{R}^2}{2(10R^2{)}^{\frac{3}{2}}}$

$B_{axis}=\frac{{\mu }_{o}I}{2\times 10\sqrt{10}R}........\left(2\right)$

Dividing equation $\left(1\right)$ to $\left(2\right)$, we get

$\frac{B_{\text{centre}}}{B_{\text{axis}}}=\frac{\frac{{\mu }_{0}I}{2R}}{\frac{{\mu }_{0}I}{2\times 10\sqrt{10}R}}$

$\frac{B_{centre}}{B_{axis}}=10\sqrt{10}$

Hence, option $\left(a\right)$ is correct.