Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Currents in dc transmission lines can be 100 A or higher. Some people areconcerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150A and a height of 8 m above the ground, what magnetic field does the line produce at ground level? Express your answer in tesla and as a percentage of the earth’s magnetic field, which is 05 G . Is this value cause for worry?

Short Answer

Expert verified

The magnetic field of earth has a magnitude of aboutB=5×10-5T , so the field produced by the transmission line is 7.5 % of the Earth’s magnetic field.

Step by step solution

01

The formula to be used in the problem

We need to find the magnetic field at the ground due to a transmission line that has current ofI=150A at a height ofr=8m above the ground. We will treat the line as a very long and straight wire.

The magnetic field due to a very long wire is given byB=μ0I2πr .

02

Find the magnetic field using the formula

B=μ0I2πrB=2×107×1508B=3.8×106T

The magnetic field of earth has a magnitude of aboutB=5×10-5T , so the field produced by the transmission line is 7.5 % of the Earth’s magnetic field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Small aircraft often have 24 V electrical systems rather than the 12 V systems in automobiles, even though the electrical power requirements are roughly the same in both applications. The explanation given by aircraft designers is that a 24 V system weighs less than a 12 V system because thinner wires can be used. Explain why this is so.

Question: A conducting sphere is placed between two charged parallel plates such as those shown in Figure. Does the electric field inside the sphere depend on precisely where between the plates the sphere is placed? What about the electric potential inside the sphere? Do the answers to these questions depend on whether or not there is a net charge on the sphere? Explain your reasoning.

Questions: A conductor that carries a net charge has a hollow, empty cavity in its interior. Does the potential vary from point to point within the material of the conductor? What about within the cavity? How does the potential inside the cavity compare to the potential within the material of the conductor?

In the circuit shown in Fig. E26.49, C = 5.90 mF, Ԑ = 28.0 V, and the emf has negligible resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2 so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after S is moved to position 2? (b) After S has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mC What is the value of the resistance R? (c) How long after S is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

Questions: When a thunderstorm is approaching, sailors at sea sometimes observe a phenomenon called “St. Elmo’s fire,” a bluish flickering light at the tips of masts. What causes this? Why does it occur at the tips of masts? Why is the effect most pronounced when the masts are wet? (Hint: Seawater is a good conductor of electricity.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free