Question

Two isolated point poles of strength $ 30{\text{ }}A{m^2} $ and $ 60{\text{ }}A{m^2} $ are placed at a distance of $ 0.3m $ . The force of repulsion is?

Answer 1

Hint :In order to find the force between the two poles, we have to use the magnetic Coulomb law for magnetic poles. The magnetic coulomb’s law is equivalent to the electrostatic coulomb law. According to the law force is given by,
 $ F = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{{{m_1}{m_2}}}{{{r^2}}} $
Where, $ m_1 $ and $ m_2 $ are pole strengths and r is the distance in between them

Complete Step By Step Answer:
As given in the question, we have the data as follows,
Pole strength of the first pole, given by m1
 $ {m_1}{\text{ }} = {\text{ }}30{\text{ }}A - {m^2} $
Pole strength of the second Pole, given by m2
 $ {m_2}{\text{ }} = {\text{ }}60{\text{ }}A - {m^2} $
Distance in between them, given by r
 $ r{\text{ }} = {\text{ }}0.3{\text{ }}m $
Also, we know that,
 $ \dfrac{{{\mu _o}}}{{4\pi }} = {10^{ - 7}} $
Now, substituting the above values in the expression of force, we will get,
 $ F = {10^{ - 7}} \times \dfrac{{30 \times 60}}{{{{(0.3)}^2}}} $
Solving the above expression will give us,
 $ F = 2 \times {10^{ - 3}}{\text{ N}} $
Hence, the Force of Repulsion between the isolated point poles is found to be $ 2 \times {10^{ - 3}}{\text{ N}} $

Additional Information:
According to the Coulomb’s law of magnetism, if there are two poles of strength m1 and m2 kept at a distance of r than the force acting in between them will be directly proportional to the product of their pole strength and inversely proportional to the square of the distance in between them.

Note :
As said in the question that these poles are similar, so it can be understood that they will be of the same nature, either the both poles will be north or the both will be south. So, the force acting in between the two poles will be repulsive in nature.