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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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Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

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The probability distribution is given as

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f(x) = 0 otherwise.

f(x) = (x/800) - (0.1)

a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

= (5.0625 - 9) - [4 - 8]

= -3.9375 + 4 = 0.0625 = 6.25%

b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

= [(120³/2400) - (0.05(120²)] - [(80³/2400) - (0.05(80²)]

= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

Variance = 11466.667 - 106.67² = 88.85

Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

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