Question

Force between two identical short bar magnets whose centres are $r$ meter apart is $4.8N$ when their axes are in the same line. If the separation is increased to $2r$ meter, the force between them is reduced to
A. $0.3N$
B. $0.6N$
C. $2.4N$
D. $1.2N$

Answer 1

Hint : If there are two magnets having dipole moment ${{M}_{1}}$ and ${{M}_{2}}$ and distance between their centre is $r$ then force between them $F=\dfrac{{{\mu }_{0}}}{4\pi }\left( \dfrac{6{{m}_{1}}{{m}_{2}}}{{{r}^{4}}} \right)$
Where ${{\mu }_{0}}=$ permeability of vacuum
${{\mu }_{0}}=4\pi \times {{10}^{-7}}{}^{N}/{}_{{{A}^{2}}}$

Complete Step By Step Answer:
Let assume magnetic moment of bar magnet $=m$
Both are identical so ${{m}_{1}}={{m}_{2}}=m$
We know that force between two bar magnet $F=\dfrac{{{\mu }_{0}}}{4\pi }\left( \dfrac{6{{m}_{1}}{{m}_{2}}}{{{r}^{4}}} \right)$
According to question when distance between bar magnet is $r$ then ${{F}_{1}}=4.8N$
Or we can say $\dfrac{{{\mu }_{0}}}{4\pi }\left( \dfrac{6{{m}_{1}},{{m}_{2}}}{{{r}^{4}}} \right)=4.8N..................\left( i \right)$
for second case when separation is increased to $2r$
${{F}_{2}}=\dfrac{{{\mu }_{0}}}{4\pi }\left( \dfrac{6{{m}_{1}},{{m}_{2}}}{{{\left( 2r \right)}^{4}}} \right)................\left( ii \right)$
Dividing equation (i) by equation (ii)
$\Rightarrow \dfrac{4.8}{{{F}_{2}}}=\dfrac{{{\left( 2 \right)}^{4}}}{1}$
$\Rightarrow {{F}_{2}}=\dfrac{4.8}{16}$
$=0.3N$
Therefore force between bar magnet when distance is $2r,{{F}_{2}}=0.3N$ .

Note :
Formula of force we used $F=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{6{{m}_{1}},{{m}_{2}}}{{{r}^{4}}}$
This formula is valid only when the distance between bar magnets is very large compared to distance between the poles of the magnet. Thus the above formula is only valid for short bar magnets or when the bar magnets are-located at for distance.