Magnetic Effects of Current and Magnetism Part 2

Given, radius of inner wire of co-axial cable $=a$Innner radius of outer shell $=b$Outer radius of outer shell $=c$Let, magnetic field at distance $x < a=B_{1}$ and magnetic field at distance$a < x < b=B_{2}$By using Ampere circuital law,$\oint B \cdot dl = n \mu_{0} l_{\text{enc}}$where, $dl =$ perimeter of small circular element,$n =$ number of turns,$\mu_{0} =$ free space permeabilityand $I =$ current.$\therefore \; B_{1} \cdot 2\pi x = n \mu_{0} I_{\text{enc}}$ $\Rightarrow \; B_{1} = \frac{\mu_{0}}{2\pi x} \frac{I_{0}}{\pi a^{2}} \cdot \pi x^{2}$$= \frac{\mu_{0}}{2\pi} \frac{I_{0} x}{a^{2}}$Now, for $B_{2}(a < x < b)$$B_{2} \cdot 2\pi x = \mu_{0} I_{0}$$B_{2} = \frac{1}{2\pi} \frac{\mu_{0} I_{0}}{x}$$\therefore$ From Eqs. (i) and (ii) we will get$\frac{B_{1}}{B_{2}} = \frac{ \frac{\mu_{0} I_{0} x}{2\pi a^{2}} }{ \frac{\mu_{0} I_{0}}{2\pi x} } = \frac{x^{2}}{a^{2}}$