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

Answer 1

Explanation:

Consider an element of thickness dr at a distance r from coildn = dr * No of turns per unit length = dr* N/(b-a) Using Biot savart law, dB = μ0Idn/2rdB = μ0INdr/(b-a)r B = integral(μ0INdr/(b-a)r) b/w a and b = μ0INln(b/a)/b-a

Answer 2

No of turns per length for a small no of turns dn and a small element of length dr =

$\frac{dn}{dr} = \frac{N}{(b-a)} \\\\dn =dr \frac{N}{(b-a)}$

Magnetic field due to a circular loop,

$dB = \frac{\mu_{0}i dn }{2r} \\\\B = \frac{\mu_{0} i N}{2(b-a)} \int\limits^b_a {\frac{1}{r} } \, dr\\\\B = \mu_{0}i\frac{N}{2(b-a)} ln\frac{b}{a}$