Ampere's Law Application Problem

[BlackMelon]

Homework Statement

I want to know why the book says that when ρ>c, there is no current and the magnetic will be zero

Homework Equations

Hø=I/2∏ρ

The Attempt at a Solution

I was trying to find the magnetic between the gap by using the formula... I've found that the magnetic magnetic field between the outer conductor and the inner one will try to cancel each other...

I think the book shouldn't just say that when ρ>c, there's no current and the field will be zero. It should say that the magnetic field from the inner and the outer conductor will completely cancel each other.

Here is the book: https://www.mediafire.com/?f0cpqvavy1ddl29

PS: The problem is on pages 190-192 of the attached book

 

[rude man]

Homework Statement

I want to know why the book says that when ρ>c, there is no current and the magnetic will be zero
[

I don't see the distinction between saying 'the fields cancel' and 'the currents cancel'. Both are correct.

 

[BlackMelon]

Rude man, if the outer layer of the conductor doesn't completely enclose the inner one but it carries the same magnitude of current as the inner one in an opposite direction, will the fields cancel each other completely? (In the case that my chosen magnetic field (H) path encloses both inner and outer conductors and vice versa)

 

[rude man]

Rude man, if the outer layer of the conductor doesn't completely enclose the inner one but it carries the same magnitude of current as the inner one in an opposite direction, will the fields cancel each other completely? (In the case that my chosen magnetic field (H) path encloses both inner and outer conductors and vice versa)

You'd have to draw me a picture of how the outer conductor somehow doesn't fully enclose the inner one. That certainly isn't the case with your illustration.

I have to warn you that the only reason you get a nice homogeneous B field around a circular path enclosing a wire carrying a current is if (1) the wire is of infinite length, and (2) the symmetry of the configuration is maintained. In other words, the configuration must be symmetrical as you go around the circular path.

 

[BlackMelon]

I've started to confuse about words from the book that I've attached to the very first post. Could you explain me the sentence at the page 191 "If the radius ρ is larger than the outer radius of the outer conductor, no current is
enclosed" and how is it different from my second post?

Thanks

 

[rude man]

I've started to confuse about words from the book that I've attached to the very first post. Could you explain me the sentence at the page 191 "If the radius ρ is larger than the outer radius of the outer conductor, no current is
enclosed" and how is it different from my second post?

Thanks

The image in your second post shows an asymmetry in the outer conductor. So the H field is not constant in magnitude as you go around your indicated path.

But the net current through the path is still zero: I + (-I) = 0 and the contour integral ∫H*ds = 0 also.

H*ds is the dot product of H and element of path length ds. H and ds are vectors.

 

[BlackMelon]

So even a broken shield in my second post can still be a perfect magnetic shield? (Just let the shield carry the same amount of current as the conductor inside?)

 

[rude man]

No. Only the integral ∫H*ds = 0. H will not be zero everywhere around a contour outside the cable. But the integral will. So, as you go around the cable there will be places where H > 0 and other places where H < 0 (opposite direction).