A 60 [Hz] high-voltagepower line carries a current of 1000 [A]. The power line is at a height of 50 [m] above the earth. What is the magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line? Since the surface of the earth is made up largely of nonmagneticmaterials, assume that the presence of the earth can be neglected in calculating the magnetic field at this frequency. Assume also that the frequency is so low so youcan neglect delay effects, so you can assume thatthe current is DC for the calculation of the magnetic field

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mtosi17 mtosi17 · Answer 1 · 2020-03-30T03:11:45+02:00

Answer:

The magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line is B=1.496x10^(-6) T.

Explanation:

The distance from the wire to a point in the surface is the heigth of the wire.

The formula for the magnetic field on any point at distance R from a wire conducting alternating current is:

[tex]B=\frac{\mu_0I}{2\pi R} =\frac{(4\pi\cdot 10^{-7}T\cdot m/A)(1000\,A)}{2\pi \cdot 50\,m} =1.496\cdot10^{-6}\,T[/tex]