Question

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 XY plane and a steady current 'I' flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

• A. $\frac{\mu_{0} NI }{2( b - a )} \ln \left(\frac{ b }{ a }\right)$
• B. $\frac{\mu_{0} NI }{2( b - a )} \ln \left(\frac{ b + a }{ b - a }\right)$
• C. $\frac{\mu_{0} NI }{2 b } \ln \left(\frac{ b }{ a }\right)$
• D. $\frac{\mu_{0} NI }{2 b } \ln \left(\frac{ b + a }{ b - a }\right)$

Answer 1

Answer: (A)

$\int\limits_{a}^{b} \frac{\mu_{0} IN }{2 r(b-a)} dr$
$=\frac{\mu_{0} I N}{2(b-a)} \int\limits_{a}^{b} \frac{d r}{r}=\frac{\mu_{0}[N}{2(b-a)} \ln \left(\frac{b}{a}\right)$