A rectangle has sides of lengths $6 c m$ and $8 c m$ . Two magnetic poles each of pole strength ' $m$ ' are placed at the ends of one side of smaller length and magnetic force acting between them is ' $F$ '. If they are moved to new positions of same rectangle so that they are diagonally opposite. Find the final force acting on each pole.

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Solution

Force $F$ action each magnet is $F = K \frac{m_{1} m_{2}}{8^{2} \times 10^{- 4}}$ ..(1) where K is constant
When magnets are diagonally opposite then final force $F^{'}$
$F^{'} = K \frac{m_{1} m_{2}}{10^{2} \times 10^{- 4}}$ ..(2)
From (1) and (2),
$F^{'} = \frac{8^{2}}{10^{2}} F = \frac{64}{100} F$

$F^{'} = \frac{16}{25} F$

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Force F action each magnet is F=Km1m282×10−4 ..(1) where K is constantWhen magnets are diagonally opposite then final force F′F′=Km1m2102×10−4 ..(2)From (1) and (2),F′=82102F=64100FF′=1625F