B field due to a wire with an alternating current

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SUMMARY

The magnetic field at a distance r from a long, thin wire carrying an alternating current I(t) = I_0 sin(ωt) can be expressed as B(r) = (μ₀ I₀ sin(ωt)) / (2πr) ẑ. This formulation is valid because the magnetic field generated by the current is directly proportional to the current itself, regardless of whether it is direct current (DC) or alternating current (AC). Inductance effects may alter the potential across the wire, but they do not affect the magnetic field calculation in this context.

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Homework Statement


I need to find the magnetic field a distance r from a long, thin wire carrying a current I(t) = I_0 \sin \omega t.

Homework Equations


Field a distance r from a wire carrying a steady current I in the z direction:

<br /> \vec B(r) = \frac{\mu_0 I}{2 \pi r} \hat \phi<br />

The Attempt at a Solution


I'm tempted to say that, in the case of the alternating current,

<br /> \vec B(r) = \frac{\mu_0 I_0 \sin \omega t}{2 \pi r} \hat \phi,<br />

but I'm not sure I'm right, and I certainly can't explain to myself why it should be Ok to assume that the DC result will be the same as the AC result. My HW problem is actually very complicated: There's a lot of stuff involving induced EMF and time-average power. But I think I'm golden on this problem if I can just figure out what the \vec B-field should be. Thanks!
 
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I think you are right. Inductance effects will cause the AC potential across the wire to be different from the DC, but you are given current and it directly causes the B field so it shouldn't matter whether DC or AC.
 

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