Energy Density of a Magnetic Field — Question 3

Section 30-10 · Physics pages 4936–4937

Question 3 Coaxial magnetic energy is distributed uniformly per logarithmic radius An ideal coaxial line carries opposite surface currents of magnitude I at radii a and b. The vacuum field between them is B = µ0I/(2πr) and vanishes elsewhere. a b azimuthal magnetic field Schematic; labels, not drawing lengths, specify the data. (a) Find energy per unit length and external inductance per unit length. (b) If b is doubled while a and I stay fixed, find the additional energy per unit length.
Show solutionHide solution
Question 3 - Solution Coaxial magnetic energy is distributed uniformly per logarithmic radius a b azimuthal magnetic field (a) Method and calculation U ℓ= ∫b a 1 2µ0 (µ0I 2πr )2 2πr dr. U/ℓ= µ0I2 4π ln(b/a), L/ℓ= µ0 2π ln(b/a) . (b) Interpretation and further calculation ∆(U/ℓ) = µ0I2 4π [ln(2b/a) −ln(b/a)]. ∆(U/ℓ) = µ0I2 4π ln 2 . Physical check The added outer annulus stores a finite energy independent of the original outer radius.

Question 3

Coaxial magnetic energy is distributed uniformly per logarithmic radius

An ideal coaxial line carries opposite surface currents of magnitude I at radii a and b. The vacuum field between them is B=μ0I/(2πr)B=\mu_0I/(2\pi r) and vanishes elsewhere.

Diagram shown in the original SVG worksheet.

  1. Find energy per unit length and external inductance per unit length.

  2. If b is doubled while a and I stay fixed, find the additional energy per unit length.

Question 3 - Solution

Coaxial magnetic energy is distributed uniformly per logarithmic radius

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Uℓ=∫ab12μ0(μ0I2πr)22πrdr.\frac U\ell=\int_a^b\frac{1}{2\mu_0}\left(\frac{\mu_0I}{2\pi r}\right)^2 2\pi r\,dr.U/ℓ=μ0I24πln⁡(b/a),L/ℓ=μ02πln⁡(b/a).\boxed{U/\ell=\frac{\mu_0I^2}{4\pi}\ln(b/a),\quad L/\ell=\frac{\mu_0}{2\pi}\ln(b/a)}.

(b) Interpretation and further calculation

Δ(U/ℓ)=μ0I24π[ln⁡(2b/a)−ln⁡(b/a)].\Delta(U/\ell)=\frac{\mu_0I^2}{4\pi}[\ln(2b/a)-\ln(b/a)].Δ(U/ℓ)=μ0I24πln⁡2.\boxed{\Delta(U/\ell)=\frac{\mu_0I^2}{4\pi}\ln2}.

Physical check

The added outer annulus stores a finite energy independent of the original outer radius.