What Produces a Magnetic Field? — Question 8

Section 28-1 · Physics pages 4486–4487

Question 8 A revolving charged ring acts as a conventional current source A thin insulating ring carries uniformly distributed charge Q=4.00 nanocoulomb and rotates at frequency f=200 revolutions per second. A fixed sensor has calibrated signed response B=kappa I for current around this ring, with kappa=.150 tesla per ampere. + + + + + + f Schematic; labels, not drawing lengths, specify the data. (a) Find the equivalent conventional current and sensor field. (b) If the charge is replaced by -Q while the mechanical rotation is unchanged, find the new current and field.
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Question 8 - Solution A revolving charged ring acts as a conventional current source + + + + + + f (a) Method and calculation T = 1/f, I = Q/T = Qf = (4.00 × 10−9)(200) = .800 µA. B = κI = (.150)(.800 × 10−6) = .120 µT. (b) Interpretation and further calculation I′ = (−Q)f = −.800 µA. B′ = κI′ = −.120 µT. Physical check Conventional current follows positive charge motion and opposes negative charge motion.

Question 8

A revolving charged ring acts as a conventional current source

A thin insulating ring carries uniformly distributed charge Q=4.00 nanocoulomb and rotates at frequency f=200 revolutions per second. A fixed sensor has calibrated signed response B=kappa I for current around this ring, with kappa=.150 tesla per ampere.

Diagram shown in the original SVG worksheet.

  1. Find the equivalent conventional current and sensor field.

  2. If the charge is replaced by -Q while the mechanical rotation is unchanged, find the new current and field.

Question 8 - Solution

A revolving charged ring acts as a conventional current source

Diagram shown in the original SVG worksheet.

(a) Method and calculation

T=1/f,I=Q/T=Qf=(4.00×10−9)(200)=.800μA.T=1/f,\quad I=Q/T=Qf=(4.00\times10^{-9})(200)=.800\,\mathrm{\mu A}.B=κI=(.150)(.800×10−6)=.120μT.B=\kappa I=(.150)(.800\times10^{-6})=.120\,\mathrm{\mu T}.

(b) Interpretation and further calculation

I′=(−Q)f=−.800μA.I\prime=(-Q)f=-.800\,\mathrm{\mu A}.B′=κI′=−.120μT.B\prime=\kappa I\prime=-.120\,\mathrm{\mu T}.

Physical check

Conventional current follows positive charge motion and opposes negative charge motion.