Question

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. When a current I passes through the coil, the magnetic field at the centre is

• A. $\frac{{\mu }_{0}NI}{b}$
  
• B. $\frac{2{\mu }_{0}NI}{a}$
  
• C. $\frac{{\mu }_{0}NI}{2(b-a)}ln\frac{b}{a}$
  
• D. $\frac{{\mu }_0{I}^N}{2(b-a)}ln\frac{b}{a}$

Answer 1

The correct option is C $\frac{{\mu }_{0}NI}{2(b-a)}ln\frac{b}{a}$


Number of turns per unit width = $\frac{N}{b-a}$
Consider an elemental ring of radius x and with thickness dx Number of turns in the ring:
$dN=\frac{Ndx}{b-a}$
Magnetic field at the centre due to the ring element
$dB=\frac{{\mu }_0\left(dN\right)i}{2x}=\frac{{\mu }_{0}i}{2}.\frac{Ndx}{(b-a)}.\frac{1}{x}$
$\therefore$ Field at the centre
$=\int dB=\frac{{\mu }_{0}Ni}{2(b-a)}{\int }_a^b\frac{dx}{x}$
=$\frac{{\mu }_{0}Ni}{2(b-a)}$ In $\frac{b}{a}$