Displacement Current — Question 2

Section 32-3 · Physics pages 5214–5215

Question 2 A smaller contour encloses only a fraction of uniform displacement current A circular vacuum capacitor of plate radius a is charged with current I. The gap field is uniform in the quasistatic approximation. Consider a centered circular contour of radius r<a. a r ⊙Ez(t) Schematic; labels, not drawing lengths, specify the data. (a) Find the displacement current enclosed by that contour and its magnetic field. (b) At r=a/3, compare enclosed current and B with their edge values.
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Question 2 - Solution A smaller contour encloses only a fraction of uniform displacement current a r ⊙Ez(t) (a) Method and calculation Jd = I/(πa2), Id,enc = Jdπr2 = Ir2/a2. Bϕ = µ0Ir/(2πa2) . (b) Interpretation and further calculation r = a/3 ⇒Id,enc = I/9. B(a/3) = B(a)/3 . Physical check A contour-area fraction determines enclosed current, but the circumference contributes another factor of radius to B.

Question 2

A smaller contour encloses only a fraction of uniform displacement current

A circular vacuum capacitor of plate radius a is charged with current I. The gap field is uniform in the quasistatic approximation. Consider a centered circular contour of radius r<a.

Diagram shown in the original SVG worksheet.

  1. Find the displacement current enclosed by that contour and its magnetic field.

  2. At r=a/3, compare enclosed current and B with their edge values.

Question 2 - Solution

A smaller contour encloses only a fraction of uniform displacement current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Jd=I/(πa2),Id,enc=Jdπr2=Ir2/a2.J_d=I/(\pi a^2),\quad I_{d,\mathrm{enc}}=J_d\pi r^2=I r^2/a^2.Bϕ=μ0Ir/(2πa2).\boxed{B_\phi=\mu_0Ir/(2\pi a^2)}.

(b) Interpretation and further calculation

r=a/3⇒Id,enc=I/9.r=a/3\Rightarrow I_{d,\mathrm{enc}}=I/9.B(a/3)=B(a)/3.\boxed{B(a/3)=B(a)/3}.

Physical check

A contour-area fraction determines enclosed current, but the circumference contributes another factor of radius to B.