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(D) The magnetic field at the centre of a circular coil is given by $B_{	ext{centre}} = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ on t

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$R \sqrt{15}$

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(D) The magnetic field at the centre of a circular coil is given by $B_{	ext{centre}} = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ on the axis of the coil is given by $B_{	ext{axis}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.According to the problem,$B_{	ext{centre}} = 64 	imes B_{	ext{axis}}$.Substituting the formulas,we get $\frac{\mu_0 I}{2R} = 64 	imes \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Canceling common terms,we have $\frac{1}{R} = \frac{64 R^2}{(R^2 + x^2)^{3/2}}$.This simplifies to $(R^2 + x^2)^{3/2} = 64 R^3$.Taking the cube root of both sides,$(R^2 + x^2)^{1/2} = 4R$.Squaring both sides,$R^2 + x^2 = 16R^2$.Therefore,$x^2 = 15R^2$,which gives $x = R\sqrt{15}$.

The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is $64$ times the magnetic field at a distance $x$ on its axis from the centre of the coil. Then,the value of $x$ is

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