A current carrying circular coil of radius 'R' produces magnetic...

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:

$B_1 = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$

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

$B_2 = \dfrac{\mu_0 I}{2R}$

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

$\dfrac{\mu_0 I}{2R} = 27 \left( \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \right)<br><br>\dfrac{1}{R} = \dfrac{27 R^2}{(R^2 + x^2)^{3/2}}<br><br>$(R² + x²)^{3/2} = 27 R³$<br><br>Taking the cube root on both sides:<br><br>$(R² + x²)^{1/2} = 3R$<br><br>Squaring both sides:<br><br>$R^2 + x² = 9R^2$<br><br>$x^2 = 8R^2$<br><br>$x = 2$\sqrt{2}$ R$<br><br>Answer: 2\sqrt{2}$\,$\text{R}$