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harkovskaia [24]
3 years ago
9

Transmission Lines and Health. Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the elec

tromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 170 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?
Physics
1 answer:
Blababa [14]3 years ago
8 0

Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:

B = μ₀I/(2πr)

B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line

Given values:

μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m

Plug in and solve for B:

B = 4π×10⁻⁷(170)/(2π(8.0))

B = 4.25×10⁻⁶T

The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:

4.25×10⁻⁶/(5.0×10⁻⁵)

= 0.085

= 8.5%

The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.

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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

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So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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