QJEEAdv11Ph7c6d: JEE Advanced 2011 Physics - Class 12 JEEAdv11Ph7c6d Solution: A long insulated copper wire is closely wound as a spiral of N…

Magnetic field at the centre of a circular loop of radius r and carrying a current $I = {{{\mu _0}I} \over {2r}}$ . The direction of the field is along z-direction if the current is anticlockwise.

Consider a small element of width dr. The current through the element is

$dI = {{total\,current\,in\,spiral} \over {total\,width\,of\,spiral}} \times width\,of\,element$

$= {{Idr} \over {(b - a)}}$

$\therefore$ $B = \int\limits_a^b {{{{\mu _0}NdI} \over {2r}} = \int\limits_a^b {{{{\mu _0}NI} \over {2(b - a)}}{{dr} \over r}} }$

$= {{{\mu _0}NI} \over {2(b - a)}}\int\limits_a^b {{{dr} \over r} = {{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)}$