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Ch 28: Sources of Magnetic Field
Chapter 28, Problem 28

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is 0.50 G. Is this value cause for worry?

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Use the Biot-Savart Law or Ampere's Law to calculate the magnetic field produced by a long straight current-carrying conductor. The formula to use is B = \( \frac{\mu_0 I}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, T \cdot m/A \)), I is the current, and r is the distance from the wire.
Substitute the given values into the formula. Here, I = 150 A and r = 8.0 m. Calculate B using the formula B = \( \frac{(4\pi \times 10^{-7} \, T \cdot m/A) \times 150 A}{2 \pi \times 8.0 m} \).
Convert the magnetic field from teslas to gauss for comparison with the Earth's magnetic field, knowing that 1 T = 10,000 G.
Calculate the percentage of the magnetic field produced by the transmission line relative to the Earth's magnetic field. Use the formula Percentage = \( \left( \frac{B_{line}}{B_{earth}} \right) \times 100\% \), where \( B_{earth} \) is 0.50 G.
Evaluate the calculated magnetic field strength in terms of health concerns by comparing it with typical exposure limits recommended by health organizations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Field Due to a Current-Carrying Wire

A current-carrying wire generates a magnetic field around it, described by Ampère's Law. The strength of this magnetic field (B) at a distance (r) from a long, straight wire can be calculated using the formula B = (μ₀ * I) / (2π * r), where μ₀ is the permeability of free space and I is the current. This principle is essential for determining the magnetic field produced by the transmission line.
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Conversion of Units

In physics, it is often necessary to convert units to ensure consistency in calculations. In this case, the magnetic field is expressed in teslas (T), while the Earth's magnetic field is given in gauss (G). The conversion factor is 1 G = 10^-4 T, which is crucial for comparing the magnetic field produced by the transmission line to the Earth's magnetic field.
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Health Effects of Electromagnetic Fields

The potential health effects of electromagnetic fields (EMF) generated by power lines have been a topic of research and debate. While some studies suggest a correlation between high EMF exposure and health risks, such as cancer, the scientific consensus indicates that the levels produced by typical power lines are generally considered safe. Understanding this context helps assess whether the calculated magnetic field poses any real health concerns.
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Related Practice
Textbook Question
A 15.0-cm-long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.
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Textbook Question
A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have?
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Textbook Question
A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (b) What total length of wire is required?
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Textbook Question
Four, long, parallel power lines each carry 100-A currents. A cross-sectional diagram of these lines is a square, 20.0 cm on each side. For each of the three cases shown in Fig. E28.25

, calculate the magnetic field at the center of the square.
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2
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Textbook Question
. Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. (a) Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other. Is the force attractive or repulsive?
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Textbook Question
. Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. (b) Each current is doubled, so that I1 becomes 10.0 A and I2 becomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20-m length of the other?
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