Answer:
A
The magnetic field strength on the ground underneath such a line is [tex]B = 2 \mu T[/tex]
B
The percentage of the earth’s magnetic field does this represent is 4%
Explanation:
Mathematically the magnetic field can be expressed as follows
[tex]B = \frac{\mu_o I}{2 \pi R}[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space [tex]= 1.2566*10^{-6} m \cdot kg \cdot s^{-2}A^{-2}[/tex]
I is the current
R is the distance
[tex]B= \frac{1.2566*10^{-6} *200 }{2* 3.142 * 20 } =2*10^{-6}T[/tex]
[tex]= 2 \mu T[/tex]
Generally the magnetic field at the Earth surface is [tex]50 \mu T[/tex]
Hence the percentage of the earth field it represents is
[tex]=\frac{2 \mu}{50 \mu} *\frac{100}{1}[/tex]
[tex]=4[/tex]%
This question involves the concepts of the magnetic field and current.
A. The magnetic field strength on the ground underneath such a line is "2 μT".
B. This represents "4.44%" of the earth's magnetic field.
A.
The magnetic field strength due to a current-carrying wire is given by the following formula:
[tex]B=\frac{\mu_oI}{2\pi r}[/tex]
where,
B = magnetic field strength = ?
I = current = 200 A
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
r = distance between wire and ground = 20 m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(200\ A)}{2\pi (20\ m)}\\\\[/tex]
B = 2 x 10⁻⁶ T = 2 μT
B.
Earth's magnetic field = 0.25 μT to 0.65 μT
Average value = 0.45 μT
Therefore,
[tex]\frac{B}{B_E}=\frac{2\ \mu T}{45\ \mu T}*100\%[/tex]
B = 4.44 % of Earth's Field
Learn more about the magnetic field here:
brainly.com/question/23096032?referrer=searchResults
