Question

Two short bar magnets with magnetic moments $400\,{\text{abAc}}{{\text{m}}^2}$ and $800\,{\text{abAc}}{{\text{m}}^2}$ are placed with their axes in the same straight line with similar poles facing each other with centres at a distance of $20\,{\text{cm}}$ from each other. Then the force of repulsion is:
A.$6\,{\text{dynes}}$
B.$12\,{\text{dynes}}$
C.$800\,{\text{dynes}}$
D.$400\,{\text{dynes}}$

Answer 1

Use the formula for force of repulsion between the two bar magnets when their like poles are facing each other. This formula gives the relation between the magnetic moments of the two bar magnets and distance between the centres of the two bar magnets.

Formula used:
The magnitude of force $F$ of repulsion between two bar magnets is given by
$F = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{6{m_1}{m_2}}}{{{r^4}}}$ …… (1)
Here, ${\mu _0}$ is permittivity of free space, ${m_1}$ is magnetic moment of first bar magnet, ${m_2}$ is magnetic moment of second bar magnet and $r$ is distance between the centres of the two bar magnet.

Complete step by step answer:
We have given that the magnetic moments of the two short bar magnets are $400\,{\text{abAc}}{{\text{m}}^2}$ and $800\,{\text{abAc}}{{\text{m}}^2}$ respectively.
${m_1} = 400\,{\text{abAc}}{{\text{m}}^2}$
$\Rightarrow{m_2} = 800\,{\text{abAc}}{{\text{m}}^2}$
The distance between the centres of the two short magnets is $20\,{\text{cm}}$.
$r = 20\,{\text{cm}}$

We have given that the two short bar magnets are placed such that the like poles of the two bar magnets are facing each other. As we know that the like poles repel each other, there will be repulsion between the like poles of two bar magnets. We have asked to determine this force of repulsion between two bar magnets.

The value of the constant $\dfrac{{{\mu _0}}}{{4\pi }}$ in the CGS system of units is 1.
$\dfrac{{{\mu _0}}}{{4\pi }} = 1$
Substitute 1 for $\dfrac{{{\mu _0}}}{{4\pi }}$, $400\,{\text{abAc}}{{\text{m}}^2}$ for ${m_1}$, $800\,{\text{abAc}}{{\text{m}}^2}$ for ${m_2}$ and $20\,{\text{cm}}$ for $r$ in equation (1).
$F = \left( 1 \right)\dfrac{{6\left( {400\,{\text{abAc}}{{\text{m}}^2}} \right)\left( {800\,{\text{abAc}}{{\text{m}}^2}} \right)}}{{{{\left( {20\,{\text{cm}}} \right)}^4}}}$
$\therefore F = 12\,{\text{dyne}}$
Therefore, the force of repulsion between two magnets is $12\,{\text{dyne}}$.

Hence, the correct option is B.

Note: There is no need to convert the units of magnetic moments and distance between the centres of the two bar magnets from CGS system of units to the SI system of units. Because the final answer for the force of repulsion between the two bar magnets is also in the CGS system of units.