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Physics / Chapter 28: Magnetic Fields / Section 28-8

Torque on a Current Loop — Question 6

Section 28-8 · Physics pages 4622–4623

Question 6 A torsion suspension makes loop deflection nonlinear in a uniform magnetic field A coil has N turns, area A and current I. At zero angular deflection its normal is perpendicular to uniform B. Deflection theta rotates the normal toward B. A torsion spring supplies restoring torque -kappa theta, with theta measured in radians. ˆn θ⃗ B coil edge Schematic; labels, not drawing lengths, specify the data. (a) Derive the equilibrium relation and solve for I from a measured theta. (b) Use N=50, A=2.00 × 10−4 square metre, B=.100 T, kappa=2.00 × 10−5 newton metre per radian and theta=30.0 degrees to find I. Test the local angular stability.
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Question 6 - Solution A torsion suspension makes loop deflection nonlinear in a uniform magnetic field ˆn θ⃗ B coil edge (a) Method and calculation τB = NIAB cos θ, κθ = NIAB cos θ. I = κθ NAB cos θ. (b) Interpretation and further calculation I = (2.00 × 10−5)(π/6) (50)(2.00 × 10−4)(.100) cos 30◦= .0120920 A. dτnet dθ = −κ −NIAB sin θ < 0 at this equilibrium. Physical check The spring angle is measured from the zero-current orientation, so the magnetic torque contains cosine here.

Question 6

A torsion suspension makes loop deflection nonlinear in a uniform magnetic field

A coil has N turns, area A and current I. At zero angular deflection its normal is perpendicular to uniform B. Deflection theta rotates the normal toward B. A torsion spring supplies restoring torque -kappa theta, with theta measured in radians.

Diagram shown in the original SVG worksheet.

  1. Derive the equilibrium relation and solve for I from a measured theta.

  2. Use N=50, A=2.00×10−42.00\times10^{-4} square metre, B=.100 T, kappa=2.00×10−52.00\times10^{-5} newton metre per radian and theta=30.0 degrees to find I. Test the local angular stability.

Question 6 - Solution

A torsion suspension makes loop deflection nonlinear in a uniform magnetic field

Diagram shown in the original SVG worksheet.

(a) Method and calculation

τB=NIABcos⁡θ,κθ=NIABcos⁡θ.\tau_B=NIAB\cos\theta,\quad\kappa\theta=NIAB\cos\theta.I=κθNABcos⁡θ.I=\frac{\kappa\theta}{NAB\cos\theta}.

(b) Interpretation and further calculation

I=(2.00×10−5)(π/6)(50)(2.00×10−4)(.100)cos⁡30∘=.0120920A.I=\frac{(2.00\times10^{-5})(\pi/6)}{(50)(2.00\times10^{-4})(.100)\cos30^\circ}=.0120920\,\mathrm A.dτnetdθ=−κ−NIABsin⁡θ<0at this equilibrium.\frac{d\tau_{\mathrm{net}}}{d\theta}=-\kappa-NIAB\sin\theta<0\quad\text{at this equilibrium}.

Physical check

The spring angle is measured from the zero-current orientation, so the magnetic torque contains cosine here.

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