A) \[\frac{{{\mu }_{0}}l}{2\pi }\]
B) \[\frac{{{\mu }_{0}}l{{r}^{2}}}{2\pi {{R}^{2}}}\]
C) \[\frac{{{\mu }_{0}}l}{2\pi r}\]
D) \[\frac{{{\mu }_{0}}lr}{2\pi {{R}^{2}}}\]
Correct Answer: D
Solution :
(d) \[\frac{{{\mu }_{0}}lr}{2\pi {{R}^{2}}}\] Let l be the current in region r < R Then, \[l'=\frac{l}{\pi {{R}^{2}}}\pi ({{r}^{2}})\] or \[l'=\frac{l{{r}^{2}}}{{{R}^{2}}}\] So, magnetic field, \[B=\frac{{{\mu }_{0}}r}{2\pi r}=\frac{{{\mu }_{0}}l{{r}^{2}}}{2\pi {{R}^{2}}r}=\frac{{{\mu }_{0}}lr}{2\pi {{R}^{2}}}\]
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