If the distance between two unlike poles is doubled then the force of attraction between them becomes ........................... times its original value.

\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is B $\frac{1}{4}$
We know that from coulomb's law of magnet-ion.
$F_{1} = \frac{μ o}{4 π} . \frac{m_{1} m_{2}}{r_{1}^{2}}$ $F_{1} = f o r c e a c t i n g b / w p o l e s μ_{o} = p e r m e a b i l i t y c o n s t a n t m_{1} & m_{2} = p o l e s t r e n g t h r_{1} = o r i g i n a l d i s t a n c e o f s e p e r a t i o n .$
Now, when the distance of sepration is doubled.
i.e, $r_{2} = 2 r_{1}$
$F_{1} = \frac{μ o}{4 π} . \frac{m_{1} m_{2}}{r_{2}^{2}}$
$\Rightarrow = \frac{μ o}{4 π} . \frac{m_{1} m_{2}}{{{(2 r}_{1})}^{2}}$
$F_{2} = \frac{F_{1}}{4}$ .
OPTION : B.

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