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(D) For two short bar magnets placed coaxially,the force of interaction $F$ is inversely proportional to the fourth power of the distance $r$ between

Vedclass · Accepted Answer

$0.3$

Vedclass · Answer

(D) For two short bar magnets placed coaxially,the force of interaction $F$ is inversely proportional to the fourth power of the distance $r$ between their centers,i.e.,$F \propto \frac{1}{r^4}$.Given the initial force $F_1 = 4.8 \ N$ at distance $r_1 = r$.When the distance is increased to $r_2 = 2r$,the new force $F_2$ is given by:$\frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^4$$\frac{F_2}{4.8} = \left( \frac{r}{2r} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$$F_2 = \frac{4.8}{16} = 0.3 \ N$.

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

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