question_answer

The magnetic field inside a solid conducting long wire at distance r from its axis is given as \[B={{B}_{0}}{{r}^{3}}\] where \[{{B}_{0}}\] is constant. Which of the following relations correctly represents current enclosed in the loop of radius r shown in the figure?

A) \[\frac{2\pi {{B}_{0}}{{r}^{3}}}{5{{\mu }_{0}}}\]     

B) \[\frac{\pi {{B}_{0}}{{r}^{4}}}{2{{\mu }_{0}}}\]

C) \[\frac{2\pi {{B}_{0}}{{r}^{2}}}{{{\mu }_{0}}}\]      

D) \[\frac{2\pi {{B}_{0}}{{r}^{4}}}{{{\mu }_{0}}}\]

Correct Answer: D

Solution :

[d] \[\oint{Bd\ell ={{\mu }_{0}}{{I}_{en}}}\] \[\oint{{{B}_{0}}{{r}^{3}}d\ell ={{\mu }_{0}}{{I}_{en}}}\] or \[{{B}_{0}}{{r}^{3}}\int\limits_{0}^{2gpr}{d\ell ={{\mu }_{0}}{{I}_{en}}}\] [Note: \[{{B}_{0}}{{r}^{3}}\] is taken out of integration because it is not varying along \[d\,\ell \] .] or \[{{B}_{0}}{{r}^{3}}(2\pi r)={{\mu }_{0}}{{I}_{en}}\] or \[{{I}_{en}}=\frac{2\pi {{B}_{0}}{{r}^{4}}}{{{\mu }_{0}}}\]