Magnitude of the magnetic field produced by a wire

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SUMMARY

The discussion focuses on calculating the magnetic field produced by a power line carrying a current of 95 A at a height of 8.5 m above the ground. The formula used is B = μ₀I / (2πr), where μ₀ is the permeability of free space, I is the current, and r is the distance from the wire to the point of measurement. The calculated magnetic field is then compared to the Earth's magnetic field, which is approximately 0.5 G. The application of Ampere's Law is emphasized for deriving the magnetic field around the wire.

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Kiyah
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Homework Statement


A power line carries a current of 95 A along the tops of 8.5 m high poles. What is the magnitude of the magnetic field produced by this wire at the ground?


How does this compare with the Earth's field of about 1/2 G?


Homework Equations



B= uo\mu/2\piI/r

The Attempt at a Solution


4\pie-7/2\pi95/8.5
 
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The formula is not appearing quite right on my screen, but you clearly have the right one. Is there any difficulty remaining?
 
Use Ampere's Law:
\oint B \bullet dl = \mu_{0}I

integrate around a closed circle of radius 8.5m centered around the wire. Since each infinitesimal point dl around the circle's circumfrence will have the same magnetic field passing through it (and because the B field is perpendicular to the circle at all points) the entire wire has the same magnitude of magnetic field passing through it.

So the integral reduces to:

B\oint dl = \mu_{0}I
B * 2 * \pi * r= \mu_{0}I
B = \frac{\mu_{0}I}{2 * \pi * r}

where r is the radius of the circle 8.5M and I is the current in the wire 95A.
 

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