Question

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 distance of x and y from the centre of the magnet is?
(A) ${2^{ - 3}}$
(B) ${2^{ - \dfrac{1}{3}}}$
(C) ${2^3}$
(D) ${2^{\dfrac{1}{3}}}$

Answer 1

Hint:
In order to solve this question, we will first find the magnetic field on the axis of the bar magnet at the given point in terms of x and then we will find the magnetic field on the equator in terms of distance y and then we will equate both values to find the ratio of their distances x and y.

Formula used:

Magnetic field due to a small bar magnet on its axis is given by ${B_{axis}} = \dfrac{{{\mu _o}2M}}{{4\pi {r^3}}}$ where M is magnetic strength of bar magnet, r is the distance at which magnetic field is to find out and ${\mu _o}$ is called relative permeability of free space.
Magnetic field on the equator point due to small bar magnet is given by ${B_{equator}} = \dfrac{{{\mu _o}M}}{{4\pi {r^3}}}$





Complete step by step solution:
According to the question, we have given that the magnetic field on the axis of the bar magnet at a distance of x is equal to the magnetic field due to bar magnetic on the equator at a distance of y. So, using formula ${B_{axis}} = \dfrac{{{\mu _o}2M}}{{4\pi {r^3}}}$ and ${B_{equator}} = \dfrac{{{\mu _o}M}}{{4\pi {r^3}}}$ we get,
$
  \dfrac{{{\mu _o}2M}}{{4\pi {x^3}}} = \dfrac{{{\mu _o}M}}{{4\pi {y^3}}} \\
  {\left( {\dfrac{x}{y}} \right)^3} = 2 \\
  \dfrac{x}{y} = {2^{\dfrac{1}{3}}} \\
 $
So, the ratio of distances x and y is ${2^{\dfrac{1}{3}}}$
Hence, the correct answer is option (D) ${2^{\dfrac{1}{3}}}$



Therefore, the correct option is D.




Note:
 It should be noted that the magnetic field due to the small bar magnet on its axis point is always twice the value of the magnetic field due to the same bar magnet at its equator point keeping the distances same on its axis and on the equator as well.