A) \[\frac{3}{2}\frac{{{h}^{2}}}{{{r}^{2}}}\]
B) \[\frac{2}{3}\frac{{{h}^{2}}}{{{r}^{2}}}\]
C) \[\frac{3}{2}\frac{{{r}^{2}}}{{{h}^{2}}}\]
D) \[\frac{2}{3}\frac{{{r}^{2}}}{{{h}^{2}}}\]
Correct Answer: A
Solution :
Field at the centre \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi in}{r}=\frac{{{\mu }_{0}}}{2}.\frac{ni}{r}\] Field at a distance h from the centre \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi ni{{r}^{2}}}{{{({{r}^{2}}+{{h}^{2}})}^{3/2}}}=\frac{{{\mu }_{0}}}{2}.\frac{ni{{r}^{2}}}{{{r}^{3}}{{\left( 1+\frac{{{h}^{2}}}{{{r}^{2}}} \right)}^{3/2}}}\] \[={{B}_{1}}{{\left( 1+\frac{{{h}^{2}}}{{{r}^{2}}} \right)}^{-3/2}}={{B}_{1}}\left( 1-\frac{3}{2}.\frac{{{h}^{2}}}{{{r}^{2}}} \right)\](By binomial theorem) Hence B2 is less than B1 by a fraction \[=\frac{3}{2}\frac{{{h}^{2}}}{{{r}^{2}}}\]
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