Given, radius of inner wire of co-axial cable
=aInnner radius of outer shell
=bOuter radius of outer shell
=cLet, magnetic field at distance
x<a=B1
and magnetic field at distance
a<x<b=B2By using Ampere circuital law,
∮B⋅dl=nµ0l‌enc ‌where,
dl= perimeter of small circular element,
n= number of turns,
µ0= free space permeability
and
I= current.
∴‌‌B1⋅2πx=nµ0I‌enc ‌ ⇒‌‌B1‌‌=‌‌⋅πx2‌‌=‌‌Now, for
B2(a<x<b)B2⋅2πx‌‌=µ0I0B2‌‌=‌‌∴ From Eqs. (i) and (ii) we will get
‌=‌=‌