Induced Electric Fields — Question 6

Section 30-5 · Physics pages 4842–4843

Question 6 A fixed-radius induction accelerator requires a relation between average and orbit fields A particle circulates at fixed radius r in an axially symmetric field. Let Bavg be the signed area-average field inside the orbit and Borb the field at the orbit. Use magnitudes in an accelerating interval: p = |q|rBorb and ˙p = |q||Eϕ|, with both relevant field magnitudes increasing. r Bavg Borb fixed circular orbit Schematic; labels, not drawing lengths, specify the data. (a) Derive the relation between their time derivatives required for fixed r. (b) Integrate it and state when the familiar factor-of-two field relation follows.
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Question 6 - Solution A fixed-radius induction accelerator requires a relation between average and orbit fields r Bavg Borb fixed circular orbit (a) Method and calculation 2πr|Eϕ| = πr2 ˙Bavg ⇒|Eϕ| = r ˙Bavg/2. ˙p = |q|r ˙Borb = |q|r ˙Bavg/2 ⇒ ˙Bavg = 2 ˙Borb . (b) Interpretation and further calculation Bavg(t) −2Borb(t) = constant . Bavg(0) = 2Borb(0) ⇒Bavg(t) = 2Borb(t) . Physical check The derivative condition alone does not set the integration constant. The initial orbit and field configuration must also be compatible.

Question 6

A fixed-radius induction accelerator requires a relation between average and orbit fields

A particle circulates at fixed radius r in an axially symmetric field. Let Bavg be the signed area-average field inside the orbit and Borb the field at the orbit. Use magnitudes in an accelerating interval: p=|q|rBorbp=|q|rB_{\mathrm{orb}} and ṗ=|q||Eϕ|\dot p=|q||E_\phi|, with both relevant field magnitudes increasing.

Diagram shown in the original SVG worksheet.

  1. Derive the relation between their time derivatives required for fixed r.

  2. Integrate it and state when the familiar factor-of-two field relation follows.

Question 6 - Solution

A fixed-radius induction accelerator requires a relation between average and orbit fields

Diagram shown in the original SVG worksheet.

(a) Method and calculation

2πr|Eϕ|=πr2Ḃavg⇒|Eϕ|=rḂavg/2.2\pi r|E_\phi|=\pi r^2\dot B_{\mathrm{avg}}\Rightarrow|E_\phi|=r\dot B_{\mathrm{avg}}/2.ṗ=|q|rḂorb=|q|rḂavg/2⇒Ḃavg=2Ḃorb.\dot p=|q|r\dot B_{\mathrm{orb}}=|q|r\dot B_{\mathrm{avg}}/2\Rightarrow\boxed{\dot B_{\mathrm{avg}}=2\dot B_{\mathrm{orb}}}.

(b) Interpretation and further calculation

Bavg(t)−2Borb(t)=constant.\boxed{B_{\mathrm{avg}}(t)-2B_{\mathrm{orb}}(t)=\text{constant}}.Bavg(0)=2Borb(0)⇒Bavg(t)=2Borb(t).B_{\mathrm{avg}}(0)=2B_{\mathrm{orb}}(0)\Rightarrow\boxed{B_{\mathrm{avg}}(t)=2B_{\mathrm{orb}}(t)}.

Physical check

The derivative condition alone does not set the integration constant. The initial orbit and field configuration must also be compatible.