Question

Repeat Exercise 28.43 for the case in which the current in the central, solid conductor is , the current in the tube is , and these currents are in the same direction rather than in opposite directions.

Short answer

(a) The magnetic field at points outside the central, solid conductor but at inside the tube is$B=\frac{{\mu }_0{I}_1}{2\pi r}$ .

(b) The magnetic field at points outside the central, solid conductor but at outside the tube is$B=\frac{{\mu }_0\left(I_1+I_2\right)}{2\pi r}$ .

Step-by-step solution

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behaving like a magnet.

Determine the magnetic field at points outside the central, solid conductor but at inside the tube and at points outside the central, solid conductor but at outside the tube

a)

Using ampere’s law to a circular path of radius:

$\begin{array}{r}\int \overrightarrow{B}\cdot d\overrightarrow{l}=BI\\ \lceil \overrightarrow{B}\cdot d\overrightarrow{l}=B\left(2\pi r\right)\end{array}$

But the enclosed current is$I_{\text{encl}}=I_1$

$\begin{array}{r}\int \overrightarrow{B}\cdot d\overrightarrow{l}={\mu }_0{l}_{\text{end}}\\ B\left(2\pi r\right)={\mu }_0{l}_1\\ B=\frac{{\mu }_0{l}_1}{2\pi r}\end{array}$

Hence, the magnetic field at points outside the central, solid conductor but at inside the tube is$B=\frac{{\mu }_0{l}_1}{2\pi r}.$ .

b)

Now the magnetic field at points outside the central, solid conductor but outside the tube is calculated as:

The enclosed current is the sum of two currents, so,

$I_{\text{end}}=I_1+I_2$

So, the magnetic field can be calculated as:

$B=\frac{{\mu }_0\left(I_1+I_2\right)}{2\pi r}$

Hence, the magnetic field at points outside the central, solid conductor but at outside the tube is$B=\frac{{\mu }_0\left(I_1+I_2\right)}{2\pi r}$ .