Q.
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 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) $\frac{\mu_{0}NI}{2b}\ln\left( \frac{b}{a} \right)$
D) $\frac{\mu_{0}NI}{2b}\ln\left( \frac{b + a}{b - a} \right)$
Explanation
(a) If we take a small strip of $dr$ at distance $r$ from centre, then number of turns in this strip would be,
$dN = \left( \frac{N}{b - a} \right)dr$
Magnetic field due to this element at the centre of the coil will be
$dB = \frac{\mu_{0}(dN)I}{2r} = \frac{\mu_{0}NI}{2(b - a)}\frac{dr}{r}$
$B = \ \int_{r = a}^{r = b}{dB} = \frac{\mu_{0}NI}{2(b - a)}\ln\left( \frac{b}{a} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.