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)  \[\frac{{{\mu }_{0}}nI}{\left( b-a \right)}{{\log }_{e}}\frac{a}{b}\]            

B)  \[\frac{{{\mu }_{0}}nI}{2\left( b-a \right)}\]

C)  \[\frac{2{{\mu }_{0}}nI}{b}\]                     

D)  \[\frac{{{\mu }_{0}}nI}{2\left( b-a \right)}{{\log }_{e}}\frac{b}{a}\]

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                 \[dB=\frac{{{\mu }_{0}}Idn}{2r}=\frac{{{\mu }_{0}}I}{2}\frac{n}{(b-a)}\frac{dr}{r}\] \[\therefore \]  \[\int_{a}^{b}{\frac{{{\mu }_{0}}Indr}{2(b-a)}=\frac{{{\mu }_{0}}In}{2(b-a)}\int_{a}^{b}{\frac{dr}{r}}}\]                 \[=\frac{{{\mu }_{0}}In}{2(b-a)}{{\log }_{e}}\left( \frac{b}{a} \right)\]