Question

(a) A DC power line for a light-rail system carries 1000 A at an angle of $30.0^\circ$ to the Earth’s $5.00 \times 10^{-5} \textrm{ T}$ field. What is the force on a 100-m section of this line? (b) Discuss practical concerns this presents, if any.

Final Answer

- $2.5 \textrm{ N}$
- $2.5 \textrm{ N}$ is insignificant. It is much less than the weight of the lines, or the force due to the wind.

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Video Transcript

This is College Physics Answers with Shaun Dychko. The force on current-carrying wire in the magnetic field is the current times the length of wire times the magnetic field strength times sine of the angle between the current in the field. So, this 1000 Amps in this light rail DC line times 100 meter length of the wire times five times ten to the negative five Tesla magnetic field strength of the Earth times sine of 30 degrees. This gives two and a half Newtons. And that's really insignificant. It's much less than with the weight of parallel lines would be and the wind blowing in the lines would also be significantly more force than this. So this force can basically be ignored.