Question

part a
use amperes law to show that the magnetic field between the conductors of a coaxial cable (see the figure (figure 1)) is (b = mu_0i/2pi r), (r) is greater than the radius of the inner wire and less than the radius of the outer cylindrical braid.
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part b

Explanation:

Step1: State Ampere's law

Ampere's law is $\oint \vec{B}\cdot d\vec{l}=\mu_0I_{enclosed}$.

Step2: Choose an Amperian loop

For a co - axial cable, choose a circular Amperian loop of radius $r$ between the conductors ($a < r< b$, where $a$ is the radius of the inner wire and $b$ is the radius of the outer cylindrical braid). The magnetic field $\vec{B}$ has a constant magnitude $B$ along the loop and is tangent to it, so $\oint \vec{B}\cdot d\vec{l}=B\oint dl$. Since $\oint dl = 2\pi r$ (circumference of the circle).

Step3: Determine the enclosed current

The current enclosed by the Amperian loop is the current $I$ in the inner solid wire, i.e., $I_{enclosed}=I$.

Step4: Solve for the magnetic field

Substituting into Ampere's law $B(2\pi r)=\mu_0I$. Then $B = \frac{\mu_0I}{2\pi r}$.

Answer:

We have shown using Ampere's law that the magnetic field between the conductors of a co - axial cable (for a radius $r$ such that the radius of the inner wire $< r<$ radius of the outer cylindrical braid) is $B=\frac{\mu_0I}{2\pi r}$ by choosing an appropriate Amperian loop, applying Ampere's law, and determining the enclosed current.