A) 2.4 N
B) 1.2N
C) 0.6 N
D) 0.3 N
Correct Answer: D
Solution :
[d] Force between two short bar magnets is given by \[F\,\,\,\,\,=\,\,\,\,\frac{{{\mu }_{0}}}{4\pi }\frac{6{{M}_{1}}{{M}_{2}}}{{{r}^{4}}}\] \[\therefore \,\,\,\,\,\,\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{r_{2}^{4}}{r_{1}^{4}}={{\left( \frac{2r}{r} \right)}^{4}}\] or \[{{F}_{2}}=\frac{{{F}_{1}}}{16}=\frac{4.8}{16}=0.3\,N\].
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