Question

The force of attraction between two unlike magnetic poles varies inversely as the square of distance between them. When the distance two poles is 10 cm. the force of attraction is 2 Dynes. Find the force of attraction between the same two poles when the distance between them is 5 cm.

Answer 1

$$\begin{gathered} \mathrm{Let}F\mathrm{denote}\mathrm{the}\mathrm{force}\mathrm{of}\mathrm{attraction}\mathrm{between}\mathrm{two}\mathrm{unlike}\mathrm{magnetic}\mathrm{poles}\mathrm{and}d\mathrm{denote}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{them}. \\ Given: \\ F\alpha \frac{1}{d^2} \\ \Rightarrow F{d}^2=k\left(k\mathrm{is}a\mathrm{constant}.\right) \\ Also, \\ 2\times {\left(10\right)}^2=\mathrm{k}\left(\because F=2d\mathrm{ynes}\mathrm{when}d=10\mathrm{cm}.\right) \\ \therefore k=200 \\ \mathrm{Thus},the\mathrm{equation}\mathrm{of}\mathrm{variation}\mathrm{is}F{d}^2=200. \\ Now, \\ F\times {\left(5\right)}^2=200\left(\because d=5\mathrm{cm}\right) \\ \therefore F=\frac{200}{25} \\ =8\mathrm{dynes} \end{gathered}$$