If the distance between two north poles of equal strength be 2 cm , the mutual force of repulsion between them becomes 2.5 dyn . What should be the distance of separation between them for which the repulsive becomes 3.6 dyn ?

Explanation

Magnetic force between two magnetic poles ,
`F=(mu_(0))/(4pi).(q_(m).q'_(m))/(r^(2))`
So if the pole - strength remains unchanged then , `Fprop(1)/(r^(2))`
Hence , for two separate distance ,
`(F_(1))/(F_(2))=((r_(2))/(r_(1)))^(2) " or, "(r_(2))/(r_(1))=sqrt((F_(1))/(F_(2)))`
`therefore" " r_(2)=r_(1)sqrt((F_(1))/(F_(2)))=2xxsqrt((2.5)/(3.6))=2xx(5)/(6)=1.67"cm"`