Torque on a Current Loop — Question 10

Section 28-8 · Physics pages 4630–4631

Question 10 Internal source resistance makes torque approach a turn-count saturation limit A coil has N turns of fixed area A, each adding resistance R1. A source of emf E and internal resistance r drives it. The field B and normal angle theta are fixed, with positive sine. Treat N as continuous for optimization. I N turns⃗ B Schematic; labels, not drawing lengths, specify the data. (a) Derive torque as a function of N and its large-N limit. (b) Find the N that gives half the limiting torque. Show that the coil heating is maximal at the same N.
Show solutionHide solution
Question 10 - Solution Internal source resistance makes torque approach a turn-count saturation limit I N turns⃗ B (a) Method and calculation I = E/(r + NR1), τ(N) = NEAB sin θ r + NR1 . τ∞= EAB sin θ R1 , τ/τ∞= NR1 r + NR1 . (b) Interpretation and further calculation τ = 1 2τ∞⇒NR1 = r ⇒N = r/R1. Pcoil = E2NR1 (r + NR1)2 , dPcoil dN = E2R1(r −NR1) (r + NR1)3 ; Pmax at NR1 = r. Physical check Increasing turns helps until source resistance becomes negligible compared with coil resistance; the ideal-voltage limit recovers turn-independent torque.

Question 10

Internal source resistance makes torque approach a turn-count saturation limit

A coil has N turns of fixed area A, each adding resistance R1. A source of emf E and internal resistance r drives it. The field B and normal angle theta are fixed, with positive sine. Treat N as continuous for optimization.

Diagram shown in the original SVG worksheet.

  1. Derive torque as a function of N and its large-N limit.

  2. Find the N that gives half the limiting torque. Show that the coil heating is maximal at the same N.

Question 10 - Solution

Internal source resistance makes torque approach a turn-count saturation limit

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=E/(r+NR1),τ(N)=NEABsin⁡θr+NR1.I=E/(r+NR_1),\quad\tau(N)=\frac{NEAB\sin\theta}{r+NR_1}.τ∞=EABsin⁡θR1,τ/τ∞=NR1r+NR1.\tau_\infty=\frac{EAB\sin\theta}{R_1},\quad\tau/\tau_\infty=\frac{NR_1}{r+NR_1}.

(b) Interpretation and further calculation

τ=12τ∞⇒NR1=r⇒N=r/R1.\tau=\tfrac12\tau_\infty\Rightarrow NR_1=r\Rightarrow N=r/R_1.Pcoil=E2NR1(r+NR1)2,dPcoildN=E2R1(r−NR1)(r+NR1)3;Pmax at NR1=r.P_{\mathrm{coil}}=\frac{E^2NR_1}{(r+NR_1)^2},\quad\frac{dP_{\mathrm{coil}}}{dN}=\frac{E^2R_1(r-NR_1)}{(r+NR_1)^3};\quad P_{\max}\text{ at }NR_1=r.

Physical check

Increasing turns helps until source resistance becomes negligible compared with coil resistance; the ideal-voltage limit recovers turn-independent torque.