Ampère's Law — Question 6

Section 29-3 · Physics pages 4702–4703

Question 6 Two infinite current sheets confine a uniform magnetic field between them Two infinite sheets parallel to xy lie at z=-a and z=+a. The lower carries surface current Kˆx and the upper carries −Kˆx, with K positive. There is no applied background field. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. Kˆx −Kˆx −a +a z Schematic; labels, not drawing lengths, specify the data. (a) Use a rectangular Ampère path and symmetry to find the field of one sheet. (b) Superpose the two fields to obtain B in the three regions.
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Question 6 - Solution Two infinite current sheets confine a uniform magnetic field between them Kˆx −Kˆx −a +a z (a) Method and calculation 2Bℓ= µ0Kℓ⇒Bsheet = µ0K/2.⃗ BKˆx(z > zs) = −µ0K 2 ˆy,⃗ BKˆx(z < zs) = +µ0K 2 ˆy. (b) Interpretation and further calculation |z| < a :⃗ B = −µ0K 2 ˆy −µ0K 2 ˆy = −µ0Kˆy .⃗ B = 0 (z > a or z < −a) . Physical check The opposite currents add their fields between the sheets and cancel outside; finite sheets would have edge fields.

Question 6

Two infinite current sheets confine a uniform magnetic field between them

Two infinite sheets parallel to xy lie at z=-a and z=+a. The lower carries surface current Kx̂K\hat x and the upper carries −Kx̂-K\hat x, with K positive. There is no applied background field. Use vacuum permeability μ0≃4π×10−7Tm/A\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A} where numerical values are needed.

Diagram shown in the original SVG worksheet.

  1. Use a rectangular Ampère path and symmetry to find the field of one sheet.

  2. Superpose the two fields to obtain B in the three regions.

Question 6 - Solution

Two infinite current sheets confine a uniform magnetic field between them

Diagram shown in the original SVG worksheet.

(a) Method and calculation

2Bℓ=μ0Kℓ⇒Bsheet=μ0K/2.2B\ell=\mu_0K\ell\Rightarrow B_{\mathrm{sheet}}=\mu_0K/2.B→Kx̂(z>zs)=−μ0K2ŷ,B→Kx̂(z<zs)=+μ0K2ŷ.\vec B_{K\hat x}(z>z_s)=-\frac{\mu_0K}{2}\hat y,\quad\vec B_{K\hat x}(z<z_s)=+\frac{\mu_0K}{2}\hat y.

(b) Interpretation and further calculation

|z|<a:B→=−μ0K2ŷ−μ0K2ŷ=−μ0Kŷ.|z|<a:\quad\vec B=-\frac{\mu_0K}{2}\hat y-\frac{\mu_0K}{2}\hat y=\boxed{-\mu_0K\hat y}.B→=0(z>a or z<−a).\boxed{\vec B=0\quad(z>a\text{ or }z<-a)}.

Physical check

The opposite currents add their fields between the sheets and cancel outside; finite sheets would have edge fields.