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A hollow wire with inner radius a and outer radius b (see below) carries uniformly distributed current I.

Use Ampere's Law to find the magnetic field created by the current for
(i) r < a
(ii) a < r < b
(iii) r > b
Draw the Ampere loops for each of the three regions

User Moosa
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Final answer:

Using Ampère's Law, the magnetic field for a hollow wire with inner radius a and outer radius b is zero for r < a, linearly increasing for a < r < b, and decreases with 1/r for r > b. Amperian loops are circular loops centered on the wire axis for each region.

Step-by-step explanation:

The question is asking to determine the magnetic field created by a hollow cylindrical wire carrying a uniform current I using Ampère's Law in three distinct regions: when r < a (inside the inner radius), a < r < b (between the inner and outer radius), and r > b (outside the outer radius). Ampère's Law, which relates the integrated magnetic field around a closed loop to the electric current passing through the loop, can be mathematically expressed as \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}.

For region (i) where r < a, there is no current enclosed by the Amperian loop, and therefore the magnetic field is zero.

For region (ii) a < r < b, the Amperian loop encloses part of the current. The enclosed current (I_{\text{enc}}) can be determined by the ratio of the area of the Amperian loop to the total area of the cross-section of the wire times the total current I. This gives us a magnetic field that varies linearly with r inside the wire, given by B = \frac{\mu_0 I r}{2\pi(a^2 - b^2)}.

For region (iii) where r > b, the entire current I is enclosed, and the magnetic field at a distance r from the center is B = \frac{\mu_0 I}{2\pi r}, which is the same as for a long straight conductor.

Amperian Loops




User Naqash Malik
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