Chapter 20, Problem 38P

To determine

The magnitude of the magnetic field at a given distance from their midpoint.

To compare: The result obtained with the Earth’s field.

Answer to Problem 38P

The magnitude of the net magnetic field $10\text{ cm}$ from their midpoint is $1.4\times {10}^{-6}\text{ T}$ . The magnetic field of the wires is about $3\%$ that of the magnetic field of the Earth.

Explanation of Solution

Given:

The current in the wires is $25\text{ A}$ at a distance of $2.8\text{ mm}$

Formula used:

The magnetic field is calculated as

  $\text{B}=\frac{{\mu }_0\text{I}}{2\pi \text{r}}$

Calculation:

Closer radius ${\text{r}}_{\text{closer}}$ will be the distance from mid-point of the nearest wire to the point and farther radius ${\text{r}}_{\text{farther}}$ will be the distance from mid-point of the farther wire to the point.

Thus, the magnetic field experienced $10\text{ cm}$ from their midpoint will be

  $\begin{array}{l}{\text{B}}_{\text{net}}=\frac{{\mu }_0\text{I}}{2\pi {\text{r}}_{\text{closer}}}-\frac{{\mu }_0\text{I}}{2\pi {\text{r}}_{\text{farther}}}\\ {\text{B}}_{\text{net}}=\frac{{\mu }_0\text{I}}{2\pi }\left(\frac{1}{{\text{r}}_{\text{closer}}}-\frac{1}{{\text{r}}_{\text{farther}}}\right)\\ {\text{B}}_{\text{net}}=\frac{(4\pi \times {10}^{-7}\text{T}·\text{m/A})(25\text{ A})}{2\pi }\left(\frac{1}{(0.10\text{ m}-0.0014\text{ m})}-\frac{1}{(0.10\text{ m}+0.0014\text{ m})}\right)\\ {\text{B}}_{\text{net}}=1.400\times {10}^{-6}\text{ T}\approx 1.4\times {10}^{-6}\text{ T}\end{array}$

On comparing the magnetic field of line with the magnetic field of the Earth of $0.5\times {10}^{-4} \text{T}$ ,

  $\frac{{\text{B}}_{\text{net}}}{{\text{B}}_{\text{Earth}}}=\frac{1.4\times {10}^{-6}\text{ T}}{0.5\times {10}^{-4}\text{ T}}=0.028$

Conclusion:

The magnitude of the net magnetic fields $10\text{ cm}$ from their midpoint is $1.4\times {10}^{-6} \text{T}$ . The magnetic field of the wires is about $3\%$ that of the magnetic field of the Earth.