Consider an element of thickness #dr# at a distance #r# from the centre of the coil.

The number of turns in this element are #dN=N/(b−a)dr#

Magnetic field* at the centre of the coil due to this element is given by the expression #dB=(μ_0dN.I)/(2r)#

Inserting values of #dN# we obtain

#dB=(μ_0N/(b−a)dr.I)/(2r)#

#=>dB=(μ_0NI)/(2(b−a)) 1/rdr#

Total magnetic field is found by Integrating from limits #a" to "b#

#B=int_a^b(μ_0NI)/(2(b−a)) 1/rdr#

#=>(μ_0NI)/(2(b−a))ln(|r|)|_a^b#, ignoring constant of integration as it is proper/definite integral

#=>(μ_0NI)/(2(b−a))(lnb-lna)#

#=>(μ_0NI)/(2(b−a))ln(b/a)#

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-

*Magnetic field at Center of a coiled wire of radius #R# carrying a current #I# can be calculated as below:

Let #dvecl# be the infinitesimal length of the coil, along the circumference, carrying current and

#dhatr# be the unit vector along the radius of the coil.

The magnetic field #dvecB# produced by this element is given by the simplified expression for Biot-Savart Law

#dvecB=(mu_0Idvecl xxdhatr)/(4piR^2)#

which simplifies as the angle #=90^@# all elements along the path and the distance to the centre is constant. The scalar part becomes

#dB=(mu_0Icdotdl )/(4piR^2)#

To find the total magnetic field we need to find the line integral of the length which is the circumference #2piR#, due to symmetry. #B=(mu_0I)/(4piR^2) oint" "dl#

#=>B=(mu_0I)/(4piR^2) 2piR#

#=>B=(mu_0I)/(2R)#

If there are #N# number of turns, the total current is #NI# and magnetic field becomes

#B=(mu_0NI)/(2R)#