Question

A bar magnet having centre O has a length of $4cm$. Point ${P_1}$​ is in the broad side-on and ${P_2}$ is in the end side-on position with \[O{P_1} = OP{2_2} = 10\;m\]. The ratio of magnetic intensities H at ${P_1}$​ and ${P_2}$​ is
(A) ${H_1}:{H_2} = 16:100$
(B) ${H_1}:{H_2} = 1:2$
(C) ${H_1}:{H_2} = 2:1$
(D) ${H_1}:{H_2} = 100:16$

Answer 1

Hint:
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 which is its end side position and then we will find the magnetic field on the equator which is its broadside and then we will find the ratio of both magnetic intensities.

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 $O{P_2} = 10m$ is ${H_2}$ then, using${B_{axis}} = \dfrac{{{\mu _o}2M}}{{4\pi {r^3}}}$ we get,
${H_2} = \dfrac{{{\mu _o}2M}}{{4\pi {{(10)}^3}}} \to (i)$
and for on Broad point which means equator, we have ${B_{equator}} = \dfrac{{{\mu _o}M}}{{4\pi {r^3}}}$ denoted by ${H_1}$ at a distance of $O{P_1} = 10m$ we get,
${H_1} = \dfrac{{{\mu _o}M}}{{4\pi {{(10)}^3}}} \to (ii)$
On dividing equation (ii) by (i) we get,
${H_1}:{H_2} = 1:2$
Hence, the correct answer is option (B) ${H_1}:{H_2} = 1:2$


Therefore, the correct option is B.




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