Quick question about Ampere's Law and how to use it?

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Ampere's Law states that the line integral of the magnetic field B around a closed loop is proportional to the enclosed current, represented as ∫B ds = μ₀(i_enc). When considering the magnetic field at the center of a loop of wire, drawing an Amperian loop inside the wire shows no enclosed current, leading to the conclusion that ∫B ds = 0, which suggests no magnetic field. However, the magnetic field at the center is non-zero due to the orthogonal relationship between B and ds, confirming that the integral evaluates to zero. The discussion also touches on the differing behaviors of magnetic fields in solid versus hollow cylindrical conductors, with solid conductors showing a linear increase in B due to the nature of current distribution, while hollow conductors exhibit a more complex, non-linear relationship. Understanding these distinctions is crucial for applying Ampere's Law correctly in various scenarios.
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I just have a quick question about how to use Ampere's Law. It says that ∫B ds = u0(i_enc), which I suppose is easy enough to understand. But I'm having trouble reconciling it with the notion of, say, the magnetic field at the center of a loop of wire.

The magnetic field at the center of a loop of wire is B = (u0 * i) / (2R). But what if I were to draw an Amperian loop inside the loop of wire? Not enclosing the loop itself, but just an arbitrary circle within the loop's boundaries. From Ampere's Law, there's no i_enc here, which makes the right side of Ampere's Law equal to 0, which implies there is no magnetic field. And yet the earlier formula says that there is a magnetic field here, and we know there is. So clearly, I'm either using Ampere's Law wrong, or it's not applicable here. I'm not sure. Help is appreciated!
 
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Jormungandr said:
there's no i_enc here, which makes the right side of Ampere's Law equal to 0, which implies there is no magnetic field.

No. it implies that ##\int {\vec B \cdot d\vec s} = 0##. What is the direction of ##\vec B## and what is the direction of ##d\vec s## along the loop that you're integrating over?
 
jtbell said:
No. it implies that ##\int {\vec B \cdot d\vec s} = 0##. What is the direction of ##\vec B## and what is the direction of ##d\vec s## along the loop that you're integrating over?

Hmm. I'm guessing since they're orthogonal, ##\int {\vec B \cdot d\vec s} = 0##, right? Okay. That makes perfect sense, thank you!

Actually, on the subject, I was wondering about something that I saw on a website somewhere. Apparently, the graph of B vs radius of a solid, cylindrical conducting material was a straight line from the origin until the surface radius and then it decreased nonlinearly according to (u0 * I) / (2*pi*r).

However, for a hollow, cylindrical conductor, the graph was 0 until the inner radius, after which time B increased nonlinearly as well until the outer radius, after which it too decreased nonlinearly. I understand why it increases nonlinearly in this case, because the i_enc term has some radius squared terms in it since the Amperian loop doesn't always enclose all of the current.

I don't, however, understand why B increases linearly for a solid material. Doesn't this also have the case where the loops don't encircle the entire current? Shouldn't this also be nonlinearly increasing, but just from the origin?
 
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