Crossed Fields: Discovery of the Electron — Question 5

Section 28-3 · Physics pages 4520–4521

Question 5 A parallel electric-field component prevents exact force cancellation at every speed Uniform electric and magnetic fields are prescribed with B nonzero. Seek a nonzero charged particle velocity for which q(⃗E +⃗v ×⃗B) = 0.⃗ B⃗ E E∥ Schematic; labels, not drawing lengths, specify the data. (a) Derive the necessary condition on E and B and find the full family of velocities when it holds. (b) If⃗E = Eyˆy + Ezˆz and⃗B = Bˆz, explain why nonzero Ez cannot be compensated by choosing a different velocity.
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Question 5 - Solution A parallel electric-field component prevents exact force cancellation at every speed⃗ B⃗ E E∥ (a) Method and calculation (⃗E +⃗v ×⃗B) ·⃗B =⃗E ·⃗B = 0.⃗ v =⃗E ×⃗B B2 + v∥ˆB when⃗E ⊥⃗B. (b) Interpretation and further calculation⃗ E ·⃗B = EzB̸ = 0 if Ez̸ = 0. F∥= qEz for every⃗v. Physical check A magnetic force is always perpendicular to B and cannot cancel electric acceleration along B.

Question 5

A parallel electric-field component prevents exact force cancellation at every speed

Uniform electric and magnetic fields are prescribed with B nonzero. Seek a nonzero charged particle velocity for which q(E→+v→×B→)=0q(\vec E+\vec v\times\vec B)=0.

Diagram shown in the original SVG worksheet.

  1. Derive the necessary condition on E and B and find the full family of velocities when it holds.

  2. If E→=Eyŷ+Ezẑ\vec E=E_y\hat y+E_z\hat z and B→=Bẑ\vec B=B\hat z, explain why nonzero Ez cannot be compensated by choosing a different velocity.

Question 5 - Solution

A parallel electric-field component prevents exact force cancellation at every speed

Diagram shown in the original SVG worksheet.

(a) Method and calculation

(E→+v→×B→)⋅B→=E→⋅B→=0.(\vec E+\vec v\times\vec B)\cdot\vec B=\vec E\cdot\vec B=0.v→=E→×B→B2+v∥B̂when E→⟂B→.\vec v=\frac{\vec E\times\vec B}{B^2}+v_\parallel\hat B\quad\text{when }\vec E\perp\vec B.

(b) Interpretation and further calculation

E→⋅B→=EzB≠0if Ez≠0.\vec E\cdot\vec B=E_zB\ne0\quad\text{if }E_z\ne0.F∥=qEzfor every v→.F_\parallel=qE_z\quad\text{for every }\vec v.

Physical check

A magnetic force is always perpendicular to B and cannot cancel electric acceleration along B.