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(D) The magnetic field on the axis of a short bar magnet at a distance $x$ is given by $B_{axis} = \frac{\mu_0}{4\pi} \frac{2M}{x^3}$.The magnetic fie

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$2^{1/3}$

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(D) The magnetic field on the axis of a short bar magnet at a distance $x$ is given by $B_{axis} = \frac{\mu_0}{4\pi} \frac{2M}{x^3}$.The magnetic field on the equatorial line of a short bar magnet at a distance $y$ is given by $B_{equator} = \frac{\mu_0}{4\pi} \frac{M}{y^3}$.Given that the magnetic fields are equal,$B_{axis} = B_{equator}$.Therefore,$\frac{\mu_0}{4\pi} \frac{2M}{x^3} = \frac{\mu_0}{4\pi} \frac{M}{y^3}$.Simplifying the equation,we get $\frac{2}{x^3} = \frac{1}{y^3}$.Rearranging for the ratio,$\frac{x^3}{y^3} = 2$.Taking the cube root on both sides,$\frac{x}{y} = 2^{1/3}$.

The magnetic field at a point $x$ on the axis of a small bar magnet is equal to the field at a point $y$ on the equator of the same magnet. The ratio of the distances of $x$ and $y$ from the centre of the magnet is

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