Question

Two equal magnetic poles placed $5cm$ 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

Answer: 0.15

Let $m$ be the pole strength of each pole.
In first case: $F=14.4\times 10^{-4}N;r=5cm=0.05m$
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 $m=\pm 6Am$
In second case: Let $r^{\prime }$ be the distance between the poles, when force of attraction becomes $F^{\prime }=1.6\times 10^{-4}N$.
Now, $F^{\prime }=\frac{{\mu }_0}{4\pi }\cdot \frac{m\times m}{r^2}$
or $1.6\times 10^{-4}=10^{-7}\times \frac{6\times 6}{r^{\prime }2}$
or $r^{\prime }=0.15m$