Question

Two short bar magnets with magnetic moments $400abAc{m^2}$ and $800abAc{m^2}$ ae placed with their axis in the same straight line similar poles facing each other with their centres at 20 cm from each other. Then the force of repulsion is:
(A) 12 dyne
(B) 6 dyne
(C) $0.08$ dyne
(D) 150 dyne

Answer 1

Hint : The force between two magnets is directly proportional to the product of their magnetic moments, and inversely proportional to the square of the distance between them. The constant of proportionality is $6 \times {10^{ - 7}}$ . We need to calculate the force firstly in Newton, then convert to dyne.

Formula used: In this solution we will be using the following formulae;
$F = \dfrac{{6\mu }}{{4\pi }}\dfrac{{{m_1}{m_2}}}{{{d^4}}}$ where $F$ is the magnitude of the force between two magnets, $\mu$ is the permeability of free space, ${m_1}$ is the magnetic moment of one magnet and ${m_2}$ is the magnetic moment of the other, $d$ is the distance between the centres.

Complete step by step answer:
To calculate the repulsion, we must recall the formula for force between two magnets, this can be given as
$F = \dfrac{{6\mu }}{{4\pi }}\dfrac{{{m_1}{m_2}}}{{{d^4}}}$ where $F$ is the force between two magnets, $\mu$ is the permeability of free space, ${m_1}$ is the magnetic moment of one magnet and ${m_2}$ is the magnetic moment of the other, $d$ is the distance between the poles.
We convert the given values to SI, and insert into the formula, hence
$F = \dfrac{{6\left( {4\pi \times {{10}^{ - 7}}} \right)}}{{4\pi }}\dfrac{{0.8 \times 0.4}}{{{{0.2}^4}}}$ (since $800ab - Ac{m^2} = 0.8A{m^2}$ and $20cm = 0.2m$ )
Thus, by computation, we have the force of repulsion to be
$F = 1.2 \times {10^{ - 4}}N$
By converting to dyne, we have
$F = 12dyne$ (since $1N = {10^5}dyne$ )
Hence the correct option is A.

Note:
For clarity, the formula $F = \dfrac{{6\mu }}{{4\pi }}\dfrac{{{m_1}{m_2}}}{{{d^4}}}$ can be proven as follows.
Generally, the magnetic field at any point due to a magnetic material with a magnetic dipole ${m_1}$ is given as
${B_1} = \dfrac{\mu }{{4\pi }}\dfrac{{2{m_1}}}{{{d^3}}}$ where $d$ is the distance of that point from the centre.
The force on another magnet of magnetic moment ${m_2}$ placed in the field is given as
${F_2} = - {m_2}\dfrac{{d{B_1}}}{{dr}}$ ( $r$ is the distance $d$ )
By differentiating, we get
$F = - {m_2}\left( {\dfrac{\mu }{{4\pi }}\dfrac{{ - 3 \times 2{m_1}}}{{{d^4}}}} \right)$
$F = \dfrac{{6\mu }}{{4\pi }}\dfrac{{{m_1}{m_2}}}{{{d^4}}}$ .