Magnitude of the magnetic field produced by a wire

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 13K views
Kiyah
Messages
1
Reaction score
0

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[tex]\mu[/tex]/2[tex]\pi[/tex]I/r

The Attempt at a Solution


4[tex]\pi[/tex]e-7/2[tex]\pi[/tex]95/8.5
 
Physics news on Phys.org
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:
[tex]\oint B \bullet dl = \mu_{0}I[/tex]

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:

[tex]B\oint dl = \mu_{0}I[/tex]
[tex]B * 2 * \pi * r= \mu_{0}I[/tex]
[tex]B = \frac{\mu_{0}I}{2 * \pi * r}[/tex]

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