Consider an element of thickness drdr at a distance rr from the centre of the coil.
The number of turns in this element are dN=N/(b−a)drdN=Nb−adr
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)
