Torque on a Current Loop — Question 4

Section 28-8 · Physics pages 4618–4619

Question 4 The best rectangular loop uses its available wire length to maximize area A single-turn rectangular loop uses a fixed wire length P and carries a maintained current I in a uniform field B. Its orientation is chosen for maximum torque. Neglect the lead length. perimeter P perimeter P Schematic; labels, not drawing lengths, specify the data. (a) Find the rectangle dimensions that maximize torque and give the maximum value. (b) Compare this value with a circular loop made from the same wire at the same current and orientation.
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Question 4 - Solution The best rectangular loop uses its available wire length to maximize area perimeter P perimeter P (a) Method and calculation 2(a + b) = P ⇒A = a(P/2 −a), dA/da = P/2 −2a. a = b = P/4, Amax = P 2/16, τsquare = IBP 2/16. (b) Interpretation and further calculation 2πr = P ⇒Acircle = P 2/(4π), τcircle = IBP 2/(4π). τcircle/τsquare = 4/π ≃1.27324. Physical check For a given perimeter, the circle encloses more area than any rectangle; the comparison assumes equal maintained current.

Question 4

The best rectangular loop uses its available wire length to maximize area

A single-turn rectangular loop uses a fixed wire length P and carries a maintained current I in a uniform field B. Its orientation is chosen for maximum torque. Neglect the lead length.

Diagram shown in the original SVG worksheet.

  1. Find the rectangle dimensions that maximize torque and give the maximum value.

  2. Compare this value with a circular loop made from the same wire at the same current and orientation.

Question 4 - Solution

The best rectangular loop uses its available wire length to maximize area

Diagram shown in the original SVG worksheet.

(a) Method and calculation

2(a+b)=P⇒A=a(P/2−a),dA/da=P/2−2a.2(a+b)=P\Rightarrow A=a(P/2-a),\quad dA/da=P/2-2a.a=b=P/4,Amax=P2/16,τsquare=IBP2/16.a=b=P/4,\quad A_{\max}=P^2/16,\quad\tau_{\mathrm{square}}=IBP^2/16.

(b) Interpretation and further calculation

2πr=P⇒Acircle=P2/(4π),τcircle=IBP2/(4π).2\pi r=P\Rightarrow A_{\mathrm{circle}}=P^2/(4\pi),\quad\tau_{\mathrm{circle}}=IBP^2/(4\pi).τcircle/τsquare=4/π≃1.27324.\tau_{\mathrm{circle}}/\tau_{\mathrm{square}}=4/\pi\simeq1.27324.

Physical check

For a given perimeter, the circle encloses more area than any rectangle; the comparison assumes equal maintained current.