Figure 29-67 shows a cross section across a diameter of a long cylindrical conductor of radius $\mathit{a}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$ carrying uniform current $\mathbf{170}\mathbf{}\mathit{A}$. What is the magnitude of the current’s magnetic field at radial distance (a) 0 (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?

• The radius of the conductor is $a=2.00cm=0.0200m$
• Current through the conductor is $i=170A$

Ampere’s law states that,

$\mathbf{\oint }\overrightarrow{\mathbf{B}}\mathbf{·}\mathit{d}\overrightarrow{\mathbf{s}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathit{i}$

The line integral in this equation is evaluated around a closed-loop called an Amperian loop. The current ion the right side is the net current encircled by the loop.

Using Ampere’s law, we can derive the magnetic field inside the wire $\mathbf{(}\mathit{r}\mathbf{}\mathbf{<}\mathbf{}\mathit{a}\mathbf{)}$ as,

$\mathit{B}\mathbf{=}\left(\frac{{\mu }_{0}i}{2\pi a^2}\right)\mathit{r}$ (i)

Here, $\mathit{i}$ is the current, $\mathit{a}$ is area of cross-section of the wire, $\mathit{r}$ is the radius.

For the first two conditions $\mathit{r}\mathbf{}\mathbf{<}\mathbf{}\mathit{a}$ , therefore, using, we can find the magnitude of the current’s magnetic field at the corresponding radii. We can also use the same equation for the third condition $\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathit{r}$. For the fourth condition, $\mathit{r}\mathbf{}\mathbf{>}\mathbf{}\mathit{a}$ , therefore, we can use equation (i) to find the corresponding magnitude of the current’s magnetic field.

The magnetic field outside the wire $\mathbf{(}\mathit{r}\mathbf{}\mathbf{>}\mathbf{}\mathit{a}\mathbf{)}$ is given by,

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$ (ii)

For $r=0$,

Since $a>r$, therefore, the magnetic field is given by,

$$\begin{gathered} B=\left(\frac{{\mu }_{0}i}{2\pi a^2}\right)r \\ =\left(\frac{{\mu }_{0}i}{2\pi a^2}\right)\times \left(0\right) \\ =0 \end{gathered}$$

The magnitude of the current’s magnetic field at a radial distance 0 cm is zero.

For $r=0.0100m$

Since,$a>r$ , therefore, from equation (i), the magnetic field is given by,

$$\begin{gathered} B=\left(\frac{{\mu }_{0}i}{2\pi a^2}\right)r \\ =\left(\frac{\left(4\pi \times {10}^{-7}T.m/A\right)\left(170A\right)}{2\pi {\left(0.0200m\right)}^2}\right)\left(0.0100m\right) \\ =8.50\times {10}^{-4}T \end{gathered}$$

The magnitude of the current’s magnetic field at a radial distance 1.00 cm is $8.50\times {10}^{-4}T$.

For $r=0.0200m$

Since,$a=r$ , therefore, from equation (ii), the magnetic field is given by,

$$\begin{gathered} B=\left(\frac{{\mu }_{0}i}{2\pi a^2}\right)r \\ =\left(\frac{\left(4\pi \times {10}^{-7}T.m/A\right)\left(170A\right)}{2\pi {\left(0.0200m\right)}^2}\right)\left(0.0200m\right) \\ =1.70\times {10}^{-3}T \end{gathered}$$

The magnitude of the current’s magnetic field at a radial distance 2.00 cm is $1.70\times {10}^{-3}T$.

For $r=0.0400m$,

Since,$a<r$ , therefore, from equation (ii), the magnetic field is given by

$$\begin{gathered} B=\left(\frac{{\mu }_{0}i}{2\pi r}\right) \\ =\left(\frac{\left(4\pi \times {10}^{-7}T.m/A\right)\left(170A\right)}{2\pi \left(0.0400m\right)}\right) \\ =8.50\times {10}^{-4}T \end{gathered}$$

The magnitude of the current’s magnetic field at a radial distance 4.00 cm is $8.50\times {10}^{-4}T$.