Chapter 20, Problem 32P

To determine

The magnitude of the magnetic field produced by the wire at the ground.

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

Answer to Problem 32P

  $B=2.2\times {10}^{-6}\text{T}$

Magnetic field is about 4% of the Earth’s magnetic field.

Explanation of Solution

Given:

Current carried by a power line is 95 A along the top of 8.5 m high poles.

Earth magnetic field is $\frac{1}{2}\text{G}$

Formula used:

The magnetic field is,

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Here, B is magnetic field, I is current, r is distance

Calculation:

Value of ${\mu }_0=4\pi \times {10}^{-7}$

Substituting the values,

  $B=\frac{4\pi \times {10}^{-7}\times 95}{2\pi \times 8.5}=\frac{190\times {10}^{-7}}{8.5}$

  $B=2.2\times {10}^{-6}\text{T}$

Given earth magnetic field is 0.50G

  $1\text{G}={10}^{-4}\text{T}$

So,

  $B_e=0.5\times {10}^{-4}\text{T}$

The ratio is,

  $\begin{array}{c}\frac{B}{B_e}=\frac{2.2\times {10}^{-6}}{0.5\times {10}^{-4}}\\ =4.4\times {10}^{-2}\end{array}$

  $\begin{array}{c}B=0.044B_e\\ \approx 4\%\text{of}{B}_e\end{array}$

Conclusion:

  $B=2.2\times {10}^{-6}\text{T}$

Magnetic field is about 4% of the Earth’s magnetic field.