Chapter 19, Problem 58QAP

To determine

An expression for the magnitude of the magnetic field in three separate regions of space; inside the inner conductor, between the two conductors, outside of the outer conductor

Answer to Problem 58QAP

  $B_{inner}=\frac{{\mu }_{0}ir}{2\pi R_i{}^2}$

  $B_{between}=\frac{{\mu }_{0}i}{2\pi r}$

  $B_{outside}=0$

Explanation of Solution

Given:

Radius of the inner conductor = R_i

Radius of outer conducting shell = R₀

Formula used:

  ${\displaystyle \sum B_{\parallel }}\Delta \mathcal{l}={\mu }_0{i}_{through}$

Calculation:

To find the field inside the inner conductor we should consider an Amperian loop of radius $r<R_i$. Inside the inner conductor the magnetic field points tangent to the Amperian loop. So $B_{\parallel }=B_{inner}$

  $\begin{array}{l}{\displaystyle \sum B_{inner}}\Delta \mathcal{l}={\mu }_0{i}_{through}\\ B_{inner}{\displaystyle \sum \Delta \mathcal{l}}={\mu }_0{i}_{through}\end{array}$

  ${\displaystyle \sum \Delta \mathcal{l}=2\pi r}$

Cross sectional area of Amperian loop = $\pi r^2$

Cross sectional area of the inner conductor = $\pi R_i{}^2$

Area of Amperian loop is less than the area of inner conductor. So current through the loop is a fraction of the total current i.

  $i_{through}=i\left(\frac{\pi r^2}{\pi R_i{}^2}\right)$

  $\begin{array}{l}{B}_{inner}(2\pi r)={\mu }_{0}i\left(\frac{\pi r^2}{\pi R_i{}^2}\right)\\ B_{inner}=\frac{{\mu }_{0}ir}{2\pi R_i{}^2}\end{array}$

To find the field between the conductors we should consider a Amperian loop of radius r, where $R_i<r<R_0$

  ${\displaystyle \sum B_{between}}\Delta \mathcal{l}={\mu }_0{i}_{through}$

  $B_{between}{\displaystyle \sum \Delta \mathcal{l}=}{\mu }_0{i}_{through}$

  ${\displaystyle \sum \Delta \mathcal{l}=2\pi r}$

The Amperian loop encloses the entire inner conductor, so $i_{through}=i$

  $\begin{array}{l}{B}_{between}(2\pi r)={\mu }_{0}i\\ B_{between}=\frac{{\mu }_{0}i}{2\pi r}\end{array}$

To find the field outside of the cable we should consider an Amperian loop with radius r, where $r>R_0$

  ${\displaystyle \sum B_{outside}}\Delta \mathcal{l}={\mu }_0{i}_{through}$

Since the current flows are in opposite directions, the net current through the loop is zero.

  $i_{through}=0$

So,

  $\begin{array}{l}{B}_{outside}(2\pi r)={\mu }_0(0)\\ B_{outside}=0\end{array}$

Conclusion:

  $B_{inner}=\frac{{\mu }_{0}ir}{2\pi R_i{}^2}$

  $B_{between}=\frac{{\mu }_{0}i}{2\pi r}$

  $B_{outside}=0$