question_answer

A coil of n number of turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current of strength\[I\] is passed through the coil, the magnetic field at its centre is

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 :

Consider an element of thickness\[dr\]at a distance\[r\]from the centre of spiral coil. Number of turns in coil\[=n\] Number of turns per unit length                 \[=\frac{n}{b-a}\] Number of turns in element\[dr=dn\] Number of turns per unit length in element\[dr\]                 \[=\frac{ndr}{b-a}\] \[ie\],    \[dn=\frac{ndr}{b-a}\] Magnetic field at its centre due to element \[dr\]is                 \[dB=\frac{{{\mu }_{0}}Idn}{2r}=\frac{{{\mu }_{0}}I}{2}\frac{n}{(b-a)}\frac{dr}{r}\] \[\therefore \]  \[B=\int_{a}^{b}{\frac{{{\mu }_{0}}Indr}{2(b-a)r}=\frac{{{\mu }_{0}}In}{2(b-a)}\int_{a}^{b}{{}}\frac{dr}{r}}\]                 \[=\frac{{{\mu }_{0}}In}{2(b-a)}{{\log }_{e}}\left( \frac{b}{a} \right)\]