Ampère's Law — Question 7

Section 29-3 · Physics pages 4704–4705

Question 7 A thick current slab has a linear interior field and constant exterior fields An infinite slab occupies −a ≤z ≤a and carries uniform volume current density J ˆx. There is no externally imposed field. The slab is infinite along x and y. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed.⃗ J z −a 0 a Schematic; labels, not drawing lengths, specify the data. (a) Derive the magnetic-field vector inside and outside the slab. (b) Find the root-mean-square field across its thickness and compare it with the surface magnitude.
Show solutionHide solution
Question 7 - Solution A thick current slab has a linear interior field and constant exterior fields⃗ J z −a 0 a (a) Method and calculation 2B(z)ℓ= µ0J(2|z|ℓ) ⇒⃗B = −µ0Jzˆy (|z| ≤a).⃗ B = −µ0Ja sgn(z)ˆy (|z| ≥a) . (b) Interpretation and further calculation B2 rms = 1 2a ∫a −a (µ0Jz)2dz = (µ0Ja)2 3 . Brms = Bsurface/ √ 3 . Physical check The signed average field across the slab is zero by antisymmetry, while the root-mean-square field is positive.

Question 7

A thick current slab has a linear interior field and constant exterior fields

An infinite slab occupies −a≤z≤a-a\le z\le a and carries uniform volume current density Jx̂J\hat x. There is no externally imposed field. The slab is infinite along x and y. 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. Derive the magnetic-field vector inside and outside the slab.

  2. Find the root-mean-square field across its thickness and compare it with the surface magnitude.

Question 7 - Solution

A thick current slab has a linear interior field and constant exterior fields

Diagram shown in the original SVG worksheet.

(a) Method and calculation

2B(z)ℓ=μ0J(2|z|ℓ)⇒B→=−μ0Jzŷ(|z|≤a).2B(z)\ell=\mu_0J(2|z|\ell)\Rightarrow\vec B=-\mu_0Jz\hat y\quad(|z|\le a).B→=−μ0Jasgn⁡(z)ŷ(|z|≥a).\boxed{\vec B=-\mu_0Ja\,\operatorname{sgn}(z)\hat y\quad(|z|\ge a)}.

(b) Interpretation and further calculation

Brms2=12a∫−aa(μ0Jz)2dz=(μ0Ja)23.B_{\mathrm{rms}}^2=\frac1{2a}\int_{-a}^a(\mu_0Jz)^2dz=\frac{(\mu_0Ja)^2}{3}.Brms=Bsurface/3.\boxed{B_{\mathrm{rms}}=B_{\mathrm{surface}}/\sqrt3}.

Physical check

The signed average field across the slab is zero by antisymmetry, while the root-mean-square field is positive.