Physics - Magnetism Question with Solution | TestHub

The magnetic field at an axial point P at a distance $x$ from the centre of a circular coil of radius $R$ is given by:

$B_1=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

The magnetic field at the centre Q of the coil is given by:

$B_2=\frac{{\mu }_{0}I}{2R}$

Given that $B_2=27B_1$, substituting the expressions for $B_1$ and $B_2$:

$\frac{{\mu }_{0}I}{2R}=27\left(\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}\right)$

$\frac{1}{R}=\frac{27R^2}{(R^2+x^2{)}^{3/2}}$

$(R^2+x^2{)}^{3/2}=27R^3$

Taking the cube root on both sides:

$(R^2+x^2{)}^{1/2}=3R$

Squaring both sides:

$R^2+x^2=9R^2$

$x^2=8R^2$

$x=2\sqrt{2}R$

Answer: $2\sqrt{2}\text{R}$