B field due to a wire with an alternating current

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The discussion focuses on determining the magnetic field around a long, thin wire carrying an alternating current described by I(t) = I_0 sin(ωt). The proposed magnetic field formula is B(r) = (μ₀ I₀ sin(ωt))/(2πr) in the φ direction. There is uncertainty about whether the DC formula can be applied directly to the AC scenario, but it is noted that the current itself generates the magnetic field. Inductance effects may complicate the situation, but the current's role in producing the B-field remains consistent. Understanding the B-field in this context is crucial for solving more complex problems involving induced EMF and time-average power.
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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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