Question

Although the evidence is weak, there has been concern in re-cent years over possible health effects from the magnetic fields generated by electric transmission lines. A typical high-voltage transmission line is$20$$m$ above the ground and carries a $200A$ current at a potential of$110KV$.

a. What is the magnetic field strength on the ground directly under such a transmission line?

b. What percentage is this of the earth’s magnetic field of$50\mu T$?

Short answer

(a) $B_{wire}=2\mu T$

(b)$4\%$

Step-by-step solution

Given Information

There has been concern in re-cent years over possible health effects from the magnetic fields

The magnetic field strength on the ground directly under such a transmission line.

The percentage is this of the earth’s magnetic field of$50\mu T$.

Given  Explanation

A current- carrying wire produces a magnetic field and the Biot-savart law enables us to calculate the magnitude and direction of this magnetic field at any point where the magnetic field due to the segment $∆\overrightarrow{s}$of current-carrying wire is given by equation $\left(29.6\right)$

${\overrightarrow{B}}_{currentsegment}=\frac{{\mu }_o}{4\pi }\frac{I∆\overrightarrow{s}\times \hat{r}}{r^2}$

The infinitesimal wire segment $∆\overrightarrow{s}$is in the same direction as the current $I,r$is the distance from $∆\overrightarrow{s}$to the point and $\hat{r}$is a unit vector that points from $\overrightarrow{dt}$to the point.

This approximation in equation $\left(1\right)$is only good if the length of the line segment is very small compared to the distance from the current element to the point. As the segment is very small compared with the distance to the center of the loop we will use the approximation form.

The direction of$\overrightarrow{dB}$is determined by applying right-hand rule to the vector product $\overrightarrow{ds}\times \hat{r}$.Hence, the magnitude of $\overrightarrow{dB}$is given by equation $\left(29.7\right)$in the form

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

Given  calculation

Now, we plug the values for $I,r$and ${\mu }_o$into equation $\left(2\right)$to get $B$of the wire

$$\begin{gathered} B_{wire}=\frac{{\mu }_o}{2\pi }\frac{I}{r} \\ =\frac{\left(4\pi \times {10}^{-7}T·m/A\right)\left(200A\right)}{2\pi \left(20m\right)} \\ =2\times {10}^{-6}T \\ =2\mu T \end{gathered}$$

Given  Simplification

$\left(b\right)$The magnetic field of the earth at the ground is $B_{earth}=50\mu T$and the line transmission applies magnetic field of $B_{wire}=2\mu T.$So, the percentage between the two magnetic fields is

$\%=\left(\frac{B_{wire}}{B_{earth}}\right)\times 100=\left(\frac{2\mu T}{50\mu T}\right)\times 100=4\%$