Chapter 20, Problem 82GP

To determine

The maximum magnetic field is produced under the power lines.

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

Answer to Problem 82GP

The magnetic field in the power line is $4\times {10}^{-5}\text{ T}$ . It is greater than the Earth’s magnetic field.

Explanation of Solution

Given:

Power lines operate at $10\text{ kV}$ and provide a maximum power of $40\text{ MW}$ .

Power lines are $3\text{ m}$ apart and are $30\text{ m}$ above the ground.

Formula used:

Magnetic field is calculated as

  $B=\frac{{\mu }_{0}I}{2r}$

Calculation:

Maximum current in the power line is

  $\begin{array}{l}P=V_{rms}{I}_{\max }\\ I_{rms}=\frac{P}{V_{rms}}\\ I_{\max }=\frac{\sqrt{2}P}{V_{rms}}\\ I_{\max }=\frac{\sqrt{2}(20\times {10}^6\text{ W})}{(10\times {10}^3\text{ V})}=2830\text{ A}\end{array}$

Considering the geometry of the figure below.

  

Distance from the ground to the power lines is

  $\begin{array}{c}r\text{ =}\sqrt{{(30\text{ m})}^2+{(1.5\text{ m})}^2}\\ \text{ =30}\text{.04 m}\end{array}$

The angle between them is

  $\begin{array}{c}\theta ={\tan }^{-1}\frac{1.5}{30}\\ =2.86°\end{array}$

The net magnetic field is

  $\begin{array}{l}{B}_{net}=B_{leftx}+B_{rightx}\\ B_{net}=2\frac{{\mu }_{0}I}{2\pi r}\cos \theta \\ B_{net}\text{= 2}\frac{(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A})(2830\text{ A})}{4\pi (30.04\text{ m)}}\cos 2.86°\\ B_{net}\approx \text{4}\times {10}^{-5}\text{ T}\end{array}$

This magnetic field is greater than the magnetic field of the Earth that is $1\times {10}^{-5}\text{ T}$ .

Conclusion:

Thus, the magnetic field in the power line is $4\times {10}^{-5}\text{ T}$ . It is greater than the Earth’s magnetic field.