A) 4 : 1 exactly
B) 4 : 1 approximately
C) 8 : 1 exactly
D) 8 : 1 approximately
Correct Answer: D
Solution :
For a magnet\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{r}^{3}}}\,\,\,\,\,\,(nearly)\] \[\Rightarrow \] \[\frac{{{B}_{2}}}{{{B}_{1}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\] \[\Rightarrow \] \[\frac{{{B}_{2}}}{{{B}_{1}}}={{\left( \frac{x}{2x} \right)}^{3}}\] \[\Rightarrow \] \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{1}{8}\] Thus,\[{{B}_{1}}:{{B}_{2}}=8:1\]approximatelyYou need to login to perform this action.
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