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 = [tex]4\pi *10^{-7} \frac{N}{A^{2}}[/tex]
By replacing in the equation ( 1 ) we have:
[tex]B = \frac{4\pi *10^{-7}N/A^{2}(180A)}{2\pi (8m)} = 4.5*10^{-6} T[/tex]
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:
[tex]\frac{B}{B_{e}} = \frac{4.5*10^{-6}}{5*10^{-5}} = 0.09 = 9%[/tex]
Thus, it is only 9% of the magnetic field of the earth.
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
