What Determines the Formula for a Magnetic Field Around a Wire?

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SUMMARY

The formula for the magnetic field around a long wire is derived using Ampère's Law, expressed as B = (μ₀/2π)(I/r). This equation illustrates that the magnetic field (B) is directly proportional to the current (I) and inversely proportional to the distance (r) from the wire. The derivation involves integrating the magnetic field around a closed loop, demonstrating the cylindrical symmetry of the field. The constant μ₀ represents the permeability of free space.

PREREQUISITES
  • Ampère's Law
  • Cylindrical coordinates
  • Magnetic field concepts
  • Basic calculus for integration
NEXT STEPS
  • Study the derivation of Ampère's Law in detail
  • Explore the implications of magnetic fields in different geometries
  • Learn about the permeability of free space (μ₀) and its significance
  • Investigate applications of magnetic fields in electrical engineering
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Physics students, electrical engineers, and anyone interested in electromagnetism and its applications in technology.

coki2000
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Hi,
Why the magnetic field's formula is B=k2i/r.Where does come from?Please prove to me.Thanks.
 
Last edited:
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coki2000 said:
Hi,
Why the magnetic field's formula is B=k2i/r.Where does come from?Please prove to me.Thanks.

There's a couple of ways to derive the equation for the magnetic field of a long wire. Ampère's Law is probably the easiest way.

Ampère's Law states

\oint _C \vec B \cdot \vec {dl} = \mu _0 I _{enc}

Where where the integral involves any arbitrary closed path (meaning the path must start and end at the same point -- in other words a loop). \mu _0 is the permeability of free space, and Ienc is the current flowing through the loop.

Consider a very long wire with current flowing through it. Now imagine tracing out a hoop around the wire, such that the wire passes through the middle of the hoop. The hoop as a radius r.

Now we can solve Ampère's Law in cylindrical coordinates. Note that in spherical coordinates, dl = rd \phi. Putting this together gives us

\int _0 ^{2 \pi} \vec B \cdot r\vec {d \phi} = \mu _0 I _{enc}

Note that r is a constant, here. Also note that the magnitude of B is also a constant due to cylindrical symmetry. So if we work with just the magnitudes, we can pull B and r out from under the integral.

Br \int _0 ^{2 \pi} d \phi = \mu _0 I _{enc}

= 2 \pi r B = \mu _0 I _{enc}

B = \frac{\mu _0}{2 \pi}\frac{I}{ r}
 
Last edited:

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