Ampère's Law — Question 2

Section 29-3 · Physics pages 4694–4695

Question 2 A uniformly conducting cylinder has different field laws inside and outside An infinite solid cylindrical wire of radius a carries total current I uniformly through its cross section toward +z. The surrounding medium is vacuum. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. r a current along +z Schematic; labels, not drawing lengths, specify the data. (a) Use a concentric circular path to derive B(r) for r less than a and r greater than a. (b) For a=2.00 mm and I=8.00 A, find the field at r=a/2, a and 2a; identify the maximum.
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Question 2 - Solution A uniformly conducting cylinder has different field laws inside and outside r a current along +z (a) Method and calculation J = I πa2 , Ienc(r) = I r2 a2 , 2πrB = µ0Ienc. B(r) = { µ0Ir/(2πa2), r ≤a, µ0I/(2πr), r ≥a.⃗ B ∥ˆϕ. (b) Interpretation and further calculation B(a) = µ0(8.00) 2π(.00200) = .800 mT. B(a/2) = .400 mT, B(2a) = .400 mT ; Bmax = B(a). Physical check The field is continuous at the surface in the absence of a singular surface-current sheet, rises linearly inside and decays outside.

Question 2

A uniformly conducting cylinder has different field laws inside and outside

An infinite solid cylindrical wire of radius a carries total current I uniformly through its cross section toward +z. The surrounding medium is vacuum. 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 concentric circular path to derive B(r) for r less than a and r greater than a.

  2. For a=2.00 mm and I=8.00 A, find the field at r=a/2, a and 2a; identify the maximum.

Question 2 - Solution

A uniformly conducting cylinder has different field laws inside and outside

Diagram shown in the original SVG worksheet.

(a) Method and calculation

J=Iπa2,Ienc(r)=Ir2a2,2πrB=μ0Ienc.J=\frac I{\pi a^2},\quad I_{\mathrm{enc}}(r)=I\frac{r^2}{a^2},\quad 2\pi rB=\mu_0I_{\mathrm{enc}}.B(r)={μ0Ir/(2πa2),r≤a,μ0I/(2πr),r≥a.B→∥ϕ̂.\boxed{B(r)=\begin{cases}\mu_0Ir/(2\pi a^2),&r\le a,\\\mu_0I/(2\pi r),&r\ge a.\end{cases}}\quad\vec B\parallel\hat\phi.

(b) Interpretation and further calculation

B(a)=μ0(8.00)2π(.00200)=.800mT.B(a)=\frac{\mu_0(8.00)}{2\pi(.00200)}=.800\,\mathrm{mT}.B(a/2)=.400mT,B(2a)=.400mT;Bmax=B(a).\boxed{B(a/2)=.400\,\mathrm{mT},\quad B(2a)=.400\,\mathrm{mT}};\quad B_{\max}=B(a).

Physical check

The field is continuous at the surface in the absence of a singular surface-current sheet, rises linearly inside and decays outside.