Answer:
a)[tex]\beta =2.46x10^{-6} T[/tex]
b) B=3.8%
Explanation:
The magnetic field can be find using the equation
[tex]\beta =\frac{u_{o}*I_{o}}{2\pi*d}[/tex]
a).
Now to determinate the current Io knowing the power and the voltage can determinate the current
[tex]P_{max}=I*V[/tex]
[tex]V=V_{rms}*\sqrt{2}[/tex]
[tex]V=240x10^3v*\sqrt{2}[/tex]
[tex]V=339.441 kV[/tex]
[tex]I_{o}=\frac{P_{max}}{V}[/tex]
[tex]I_{o}=\frac{46x10^6W}{339.411x10^3v}[/tex]
[tex]I_{o}=135.5 A[/tex]
Replacing
[tex]u_{o}=4\pi x10^{-7}\frac{T*m}{A}[/tex]
[tex]d=11m[/tex]
[tex]\beta =\frac{4\pix10^{-7}\frac{T*m}{A} *135.5A}{2\pi*11m}[/tex]
[tex]\beta =2.46x10^{-6} T[/tex]
b).
The magnetic field of the earth is about 68uT so compare with this field
%[tex]=\frac{\beta_{wire}}{\beta_{earth}}[/tex]
%[tex]=\frac{2.46uT}{64uT}*100[/tex]
%=3.8 %
A) The estimated magnetic field experienced under such power line; 2.46 * 10⁻⁶T
B) when compared to the earth's magnetic field = 3.8%
Given Data :
Height of transmission wire ( d ) = 11m
voltage of power line ( Vrms ) = 240 kV
Max power in Power line ( Pmax ) = 46 Mw
u₀ = 410⁻⁷ T.m / A
a) Determine the magnetic field of the power line
β = [tex](u_{0}*I_{0} )/ 2\pi d[/tex] ---- ( 1 )
where; I₀ = Pmax / V ---- ( 2 )
and V = Vrms * √2 ---- ( 3 )
Back to equation ( 3 )
V = 240 * 10³ * √2
= 339.441 kV
Back to equation ( 2 )
I₀ = ( 46 * 10⁶ ) / ( 339.441 * 10³ )
= 135.5 A
Insert given and calculated values into equation ( 1 )
β = ( 410⁻⁷ * 135.5 ) / (2 * π * 11 )
= 2.46 * 10⁻⁶ T
B) The ratio of magnetic field produced by wire to Earth's magnetic field
Bwire / Bearth
= 2.46 * 10⁻⁶ / 64 * 10⁻⁶
= 0.038 = 3.8%
Hence we can conclude that The estimated magnetic field experienced under such power line; 2.46 * 10⁻⁶T, when compared to the earth's magnetic field = 3.8%
Learn more : https://brainly.com/question/22403676
