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. $\frac{{\mu }_{0}IN}{2(b-a)}ln\frac{b}{a}$
• B. $\frac{{\mu }_{0}IN}{2(b-a)}ln\left(\frac{b+a}{b-a}\right)$
• C. $\frac{{\mu }_{0}IN}{2(b-a)}ln\frac{a}{b}$
• D. $\frac{{\mu }_{0}IN}{2\left(b\right)}ln\left(\frac{b+a}{b-a}\right)$

Answer 1

The correct option is A

$\frac{{\mu }_{0}IN}{2(b-a)}ln\frac{b}{a}$

$B=\int dB={\int }_a^b\frac{{\mu }_0\left(dN\right)I}{2x}=\frac{{\mu }_{0}I}{2}{\int }_a^b\frac{Ndx}{(b-a)x}$

$=\frac{{\mu }_{0}IN}{2(b-a)}ln\frac{b}{a}$