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The magnetic field due to a short magnet at a point on its axis at distance X cm from the middle point of the magnet is $200gauss$. The magnetic field at a point on the neutral axis at a distance X cm from the middle of the magnet is (A) $100Gauss$(B) $400Gauss$(C) $50Gauss$(D) $200Gauss$

Last updated date: 18th Jul 2024
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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 and then we will find the magnetic field on the equator which is also the neutral axis.

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 magnetic field on the axis of bar magnet at a distance of $Xcm$ then and$B = 200Gauss$, using${B_{axis}} = \dfrac{{{\mu _o}2M}}{{4\pi {r^3}}}$ we get,
$200 = \dfrac{{{\mu _o}2M}}{{4\pi {{(X)}^3}}} \to (i)$ and the point on equator at a distance of $Xcm$ we have ${B_{equator}} = \dfrac{{{\mu _o}M}}{{4\pi {r^3}}}$ so we get,
${B_{equator}} = \dfrac{{{\mu _o}M}}{{4\pi {{(X)}^3}}} \to (ii)$
Divide the equation (i) and (ii) we get,
$\dfrac{{200}}{{{B_{equator}}}} = 2 \\ {B_{equator}} = 100Gauss \\$
Hence, the correct answer is option (A) $100Gauss$

Therefore, the correct option is A.

Note:
It should be noted that the magnetic field due to a small bar magnet on its axis point which is the end 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.