Question

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is-

• A. $\frac { \mu _ { 0 } \mathrm { NI } } { \mathrm { b } }$
• B. $\frac { 2 \mu _ { 0 } \mathrm { NI } } { \mathrm { a } }$
• C. $\frac { \mu _ { 0 } \mathrm { NI } } { 2 ( \mathrm {~b} - \mathrm { a } ) }$ log $\frac { \mathrm { b } } { \mathrm { a } }$
• D. $\frac { \mu _ { 0 } \mathrm { IN } } { 2 ( \mathrm {~b} - \mathrm { a } ) }$ log $\frac { \mathrm { a } } { \mathrm { b } }$

Answer 1

Answer: (C)

$\frac { \mu _ { 0 } \mathrm { NI } } { 2 ( \mathrm {~b} - \mathrm { a } ) }$ log $\frac { \mathrm { b } } { \mathrm { a } }$

Answer 2

An element is assumed at distance x from center whose width is dx. No. of turns in width b-a = N

\f0 No. of turns is width dx = n =

dB = $\frac { \mu _ { 0 } \mathrm { ni } } { 2 \mathrm { x } }$ = $\frac { \mu _ { 0 } \mathrm { iN } } { 2 ( b - a ) } \frac { d x } { x }$

B = $\frac { \mu _ { 0 } \mathrm { i } N } { 2 ( b - a ) } \int _ { a } ^ { b } \frac { d x } { x }$ =