This question asks for the specific distance on the axis of a circular coil where the magnetic field strength is reduced to one twenty-seventh (1/27th) of its value at the center.

Calculating Magnetic Field on Coil Axis

The magnetic field ($B_x$) on the axis of a circular coil with radius $R$, carrying current $I$, at a distance $x$ from the center is given by the formula:

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

The magnetic field ($B_0$) at the center of the coil (where $x=0$) is:

$ B_0 = \frac{\mu_0 I}{2R} $

Determining the Required Distance

We are given the condition that the magnetic field on the axis is $\frac{1}{27}$th of the field at the center:

$ B_x = \frac{1}{27} B_0 $

Substituting the formulas:

$ \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} = \frac{1}{27} \left( \frac{\mu_0 I}{2R} \right) $

Simplify by cancelling common terms ($\mu_0 I / 2$):

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

Rearrange the equation:

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

To solve for $x$, raise both sides to the power of $\frac{2}{3}$:

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

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

$ 3^2 R^2 = R^2 + x^2 $

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

Now, isolate $x^2$:

$ x^2 = 9 R^2 - R^2 $

$ x^2 = 8 R^2 $

Finally, take the square root to find $x$:

$ x = \sqrt{8 R^2} $

$ x = \sqrt{8} R $

$ x = 2\sqrt{2} R $

Therefore, the distance from the center on the axis where the magnetic induction is $\frac{1}{27}$th of the value at the center is $2\sqrt{2} R$. This corresponds to Option A.