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Vilka [71]
4 years ago
14

Currents in dc transmission lines can be 100 A or more. Some people have expressed concern that the electromagnetic fields (EMFs

) from such lines near their homes could cause health dangers.
Part A

For a line with current 180 A and at a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.

B=____T

Part B

What magnetic field does the line produce at ground level as a percent of the earth's magnetic field, which is 0.50 gauss.

B/Be=____%
Physics
2 answers:
S_A_V [24]4 years ago
8 0

Explanation:

This problem can be solved by using the equation

B = (μ0*I)/(2π*r)        ( 1 )

where μ0 is the magnetic permeability, I is the current in the wire and  r is the distance to the wire, where the magnetic field B is calculated. The equation (1) says that a current in a wire produce a magnetic field in a distance r from the wire.

Part A

Taking into account that μ0 = 4\pi *10^{-7} \frac{N}{A^{2}}

By replacing in the equation ( 1 ) we have:

B = \frac{4\pi *10^{-7}N/A^{2}(180A)}{2\pi (8m)} = 4.5*10^{-6} T

Part B

In tis case is only necessary to calculate B/Be (Be is the magnetic field of the earth), but taking into account that tha magnetic field of the earth must be in teslas.

1 gauss = 10^(-4) T

0.50 gauss = 0.50*(10^(-4)) T = 5*10^(-5) T

Hence, the proportion between B and Be:

\frac{B}{B_{e}} = \frac{4.5*10^{-6}}{5*10^{-5}} = 0.09 = 9%

Thus, it is only 9% of the magnetic field of the earth.

GuDViN [60]4 years ago
8 0

Answer:

Explanation:

Current in distribution line

I=100A

EMF at such height from the earth can cause hazard

a. To know how strong the magnetic field are at a height of 8m above the earth

And the line current is 180A

I=180A

Magnetic field is given as

B= μoI /2πa

μo = permeability of free space

I = magnitude of current in the wire

a = perpendicular distance between point P and the straight current carrying conductor.

a=8m from earth

μo is a constant and it value is

μo=4π×10^-7 H/m

Then,

B=μoI /2πa

B= 4π×10^-7 × 180 / (2π ×8)

B= 4.5×10^-6 T

b. Earth magnetic flux is 0.5gauss

Note, 1 gauss = 10^-4 Telsa

Therefore, the magnetic field on earth becomes

Be=0.5gauss = 0.5 × 10^-4

Be=5×10^-5

Then, the percentage of magnetic field due to magnetic on earth is

B/Be ×100

(4.5×10^-6) / (5×10^-5) ×100

0.09×100

9%

The magnetic field produce a 9% percentage on the ground level

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