Question

A coaxial cable consists of a solid inner conductor of radius a =0.2 m carrying total current of 5A, surrounded by a concentric cylindrical tube of inner radius b=0.4 m and outer radius c=0.6m carrying current of 5A. The conductors carry equal and opposite currents I distributed uniformly across their cross-sections. Determine the magnetic field at a distance r=0.5m from the axis.  (10p)

Answer 1

Gives

a=0.2 m

Current=5A

b=0.4m

c=0.6m

r=0.5m

We know that

Ampear law in closed loop

$\oint B .dl=\mu_0I_{enclosed}$

$I_{en}=I_0$

$\oint B.dl=\mu_0(I-I_0)$

$\oint dl=2\pi r$

Area $\pi(c^2-b^2)$ Flow current=$I$

Area 1m²flow current$=\frac{I}{\pi(c^2-b^2)}$

Area $\pi(r^2-b^2)$ Flow current$(I_0)$ $=\frac{I}{\pi(c^2-b^2)}\times\pi(r^2-b^2)$

$I_0=\frac{\pi (r^2-b^2)}{\pi(c^2-b^2)}I$

$I_0=\frac{(r^2-b^2)}{(c^2-b^2)}I$

$B=\frac{\mu_0I}{2\pi r}\times(1-\frac{r^2-b^2}{c^2-b^2})$

Put value

Where

$\mu_0=4\pi\times10^{-7} Wb/A -m$

$B=\frac{4\times3.14\times10^{-7}\times5}{2\times3.14\times 0.5}\times(1-\frac{0.5^2-0.4^2}{0.6^2-0.4^2})$

$$B=2\times10^{-6}\times0.55=1.1\times10^{-6}T$$