Question

Force between two identical short bar magnets whose centres are $r$ metre apart is $4.8N$ when their axes are in the same line. If the separation is increased to $2r$ metre, the force between them is reduced to?

Answer 1

Step 1: Given data

Two bar magnets placed axially at a distance $r$.

The force between them is $4.8N$.

The separation is increased to $2r$ metre.

Step 2: Formula used

Force between two magnetic dipoles is $F=\frac{{\mu }_0}{4\mathrm{\pi }}\left(\frac{6M_1{M}_2}{d^4}\right)$

where ${\mu }_0$ is the permeability of free space, $M_1$ and $M_2$ are the dipole moments of the two magnets, and $d$ is the distance of separation between them.

From this equation, we can see that $F\propto \frac{1}{d^4}\propto \frac{1}{r^4}$

Step 2: Calculate the Required Force

When the separation is increased to $2r$, $F\text{'}\propto \frac{1}{{\left(2r\right)}^4}$

Taking the ratio of the two forces we get,

$\frac{F\text{'}}{F}=\frac{r^4}{{\left(2r\right)}^4}$

$\Rightarrow F\text{'}=\frac{F}{2^4}$

$=\frac{4.8}{16}=0.3N$

Therefore, force is reduced to $0.3N$ between the two identical short bar magnets.