A) \[\frac{{{\mu }_{0}}nI}{(b-a)}{{\log }_{e}}\frac{a}{b}\]
B) \[\frac{{{\mu }_{0}}nI}{2(b-a)}\]
C) \[\frac{2{{\mu }_{0}}nI}{b}\]
D) \[\frac{{{\mu }_{0}}nI}{2(b-a)}{{\log }_{e}}\frac{b}{a}\]
Correct Answer: D
Solution :
Let us consider an element of thickness dr at a distance r from the centre. Number of turns per unit length\[=\frac{n}{b-a}\] Number of turns per unit length in element dr; \[dn=\frac{n\cdot dr}{b-a}\] Magnetic field at the center due to element dr is \[dB=\frac{{{\mu }_{0}}I}{2r}.\,dn=\frac{{{\mu }_{0}}I}{2}\cdot \frac{n\,dr}{(b-a)r}\] \[\therefore \] \[B=\int\limits_{a}^{b}{\frac{{{\mu }_{0}}I\,ndr}{2(b-a)\,\cdot r}}=\frac{{{\mu }_{0}}I\,n}{2\,(b-a)}\int\limits_{a}^{b}{\frac{dr}{r}}\] \[=\frac{{{\mu }_{0}}I\,n}{2(b-a)}{{\log }_{e}}\left( \frac{b}{a} \right).\]
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