A) \[B=\frac{{{\mu }_{0}}\pi C}{4{{r}^{2}}}({{r}^{2}}-{{a}^{2}})\]
B) \[B=\frac{{{\mu }_{0}}C}{4{{r}^{2}}}({{r}^{4}}-{{a}^{4}})\]
C) \[B=\frac{{{\mu }_{0}}\pi C}{4r}({{r}^{2}}-{{a}^{2}})\]
D) \[B=\frac{{{\mu }_{0}}C}{4r}({{r}^{4}}-{{a}^{4}})\]
Correct Answer: D
Solution :
[d]: Given : Current density\[J=C{{r}^{2}}\] For a region a < r < b. The Amperian loop is a circle labelled as 1 According to Amperes circuital law \[=\oint{{\vec{B}}}.d\vec{l}={{\mu }_{0}}I\] \[B(2\pi r)={{\mu }_{0}}\int_{{}}^{{}}{JdA}={{\mu }_{0}}\int\limits_{a}^{r}{Cr{{'}^{2}}}(2\pi r'dr')\] \[={{\mu }_{0}}C2\pi \int\limits_{a}^{r}{r{{'}^{3}}dr'}={{\mu }_{0}}C2\pi \left[ \frac{r{{'}^{4}}}{4} \right]_{a}^{r}\] \[B2\pi r=\frac{{{\mu }_{0}}C2\pi }{4}[{{r}^{4}}-{{a}^{4}}];B=\frac{{{\mu }_{0}}C}{4r}[{{r}^{4}}-{{a}^{4}}]\]You need to login to perform this action.
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