Question

$A$ long insulated copper wire is closely wound as a spiral of '$N$' turns. The spiral has inner radius ` $a$' and outer radius ${}^{\prime }{b}^{\prime }$. 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 :

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

Answer 1

Answer: (A)

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

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 $dN=\frac{N}{b-a}dr$
Thus magnetic field at the centre O due to this ring $dB=\frac{{\mu }_{o}IdN}{2r}$
We get $dB=\frac{{\mu }_{o}INdr}{2(b-a)r}$
Net magnetic field at centre of spiral $B={\int }_a^b\frac{{\mu }_{o}IN}{2(b-a)}\frac{dr}{r}$
$\therefore$ $B=\frac{{\mu }_{o}IN}{2(b-a)}{\int }_a^b\frac{dr}{r}$
Or $B=\frac{{\mu }_{o}IN}{2(b-a)}\times \ln r{|}_a^b$
Or $B=\frac{{\mu }_{o}IN}{2(b-a)}\ln \frac{b}{a}$