A long insulated copper wire is closely wound as a spiral of 'N' turns. The spiral has inner radius 'a' and outer radius 'b'. The spiral lies in the X-Y plane and a steady current 'I' flows through the wire. The Z-component of the magnetic field at the center of the spiral is ?

Let us consider an elementary ring of radius 'r' and thickness 'dr' in which current 'I' is flowing. Number of turns in this elementary ring dN = N/(b-a) dr Thus magnetic field at the centre O due to this ring dB = µ0IdN/2r We get dB = µ0INdr/2(b-a)r Net magnetic field at centre of spiral B = ∫ab µ0IN/2(b-a) dr/r ∴ B = µ0IN/2(b-a) ∫ab dr/r Or B = µ0IN/2(b-a) × ln(r)|ab Or B = µ0IN/2(b-a) ln(b/a)