Question

Two equal magnetic poles placed $5\, cm$ in air attract each other with a force of $14.4 \times 10^{-4} \,N$. How far (in $m$ ) from each other should they be placed so that the force of attraction will be $1.6 \times 10^{-4} \,N$ ?

Answer 1

Let $m$ be the pole strength of each pole.
In first case: $F=14.4 \times 10^{-4} N ; r=5 \,cm =0.05 \,m$
Now, $ F=\frac{\mu_{0}}{4 \pi} \cdot \frac{m \times m}{r^{2}}$
or $ 14.4 \times 10^{-4}=10^{-7} \times \frac{m^{2}}{(0.05)^{2}}$
or $m=\pm 6\, Am$
In second case: Let $r'$ be the distance between the poles, when force of attraction becomes $F'=1.6 \times 10^{-4} N$.
Now, $ F'=\frac{\mu_{0}}{4 \pi} \cdot \frac{m \times m}{r^{2}}$
or $ 1.6 \times 10^{-4}=10^{-7} \times \frac{6 \times 6}{r'2}$
or $ r'=0.15 \,m$