The Definition of B — Question 9

Section 28-2 · Physics pages 4508–4509

Question 9 An acceleration component along the velocity reveals a nonmagnetic force A particle has positive charge-to-mass ratio q/m = 2.00 × 107 C/kg, velocity 4.00 × 104ˆx m/s and acceleration (−2ˆx + 3ˆy)106 m/s squared. x y⃗ v⃗ a Schematic; labels, not drawing lengths, specify the data. (a) Explain quantitatively why a magnetic field alone cannot produce this acceleration. (b) Assume the electric field has only an x component and the magnetic field only a z component. Determine both fields.
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Question 9 - Solution An acceleration component along the velocity reveals a nonmagnetic force x y⃗ v⃗ a (a) Method and calculation⃗ a ·⃗v = (−2 × 106)(4 × 104) = −8.00 × 1010 m2/s3̸ = 0.⃗ aB ·⃗v = (q/m)(⃗v ×⃗B) ·⃗v = 0. (b) Interpretation and further calculation ax = (q/m)Ex ⇒Ex = −.100 V/m. ay = −(q/m)vBz ⇒Bz = −3.75 × 10−6 T. Physical check Only the electric component changes the speed at this instant under the stated field model.

Question 9

An acceleration component along the velocity reveals a nonmagnetic force

A particle has positive charge-to-mass ratio q/m=2.00×107q/m=2.00\times10^7 C/kg, velocity 4.00×104x̂4.00\times10^4\hat x m/s and acceleration (−2x̂+3ŷ)106(-2\hat x+3\hat y)10^6 m/s squared.

Diagram shown in the original SVG worksheet.

  1. Explain quantitatively why a magnetic field alone cannot produce this acceleration.

  2. Assume the electric field has only an x component and the magnetic field only a z component. Determine both fields.

Question 9 - Solution

An acceleration component along the velocity reveals a nonmagnetic force

Diagram shown in the original SVG worksheet.

(a) Method and calculation

a→⋅v→=(−2×106)(4×104)=−8.00×1010m2/s3≠0.\vec a\cdot\vec v=(-2\times10^6)(4\times10^4)=-8.00\times10^{10}\,\mathrm{m^2/s^3}\ne0.a→B⋅v→=(q/m)(v→×B→)⋅v→=0.\vec a_B\cdot\vec v=(q/m)(\vec v\times\vec B)\cdot\vec v=0.

(b) Interpretation and further calculation

ax=(q/m)Ex⇒Ex=−.100V/m.a_x=(q/m)E_x\Rightarrow E_x=-.100\,\mathrm{V/m}.ay=−(q/m)vBz⇒Bz=−3.75×10−6T.a_y=-(q/m)vB_z\Rightarrow B_z=-3.75\times10^{-6}\,\mathrm T.

Physical check

Only the electric component changes the speed at this instant under the stated field model.