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