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Physics / Chapter 28: Magnetic Fields / Section 28-3

Crossed Fields: Discovery of the Electron — Question 1

Section 28-3 · Physics pages 4512–4513

Question 1 Opposing electric and magnetic forces select a speed independently of charge sign A beam enters along +x between plates with⃗E = −2.40 × 104ˆy V/m and⃗B = −8.00 × 10−3ˆz T. Ignore gravity and collisions. positive plate negative plate⃗ v⃗ E Schematic; labels, not drawing lengths, specify the data. (a) Find the speed that produces a straight trajectory. (b) For an electron at this speed, give the directions and magnitudes of the two forces, using e = 1.602 × 10−19 C.
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Question 1 - Solution Opposing electric and magnetic forces select a speed independently of charge sign positive plate negative plate⃗ v⃗ E (a) Method and calculation⃗ v ×⃗B = vBˆy, Fy = q(−E + vB) = 0. v0 = E/B = 3.00 × 106 m/s. (b) Interpretation and further calculation⃗ FE = +eEˆy = 3.8448 × 10−15ˆy N.⃗ FB = −ev0Bˆy = −3.8448 × 10−15ˆy N. Physical check Both forces reverse for the opposite charge sign, so the selected speed remains E/B.

Question 1

Opposing electric and magnetic forces select a speed independently of charge sign

A beam enters along +x between plates with E→=−2.40×104ŷ\vec E=-2.40\times10^4\hat y V/m and B→=−8.00×10−3ẑ\vec B=-8.00\times10^{-3}\hat z T. Ignore gravity and collisions.

Diagram shown in the original SVG worksheet.

  1. Find the speed that produces a straight trajectory.

  2. For an electron at this speed, give the directions and magnitudes of the two forces, using e=1.602×10−19e=1.602\times10^{-19} C.

Question 1 - Solution

Opposing electric and magnetic forces select a speed independently of charge sign

Diagram shown in the original SVG worksheet.

(a) Method and calculation

v→×B→=vBŷ,Fy=q(−E+vB)=0.\vec v\times\vec B=vB\hat y,\quad F_y=q(-E+vB)=0.v0=E/B=3.00×106m/s.v_0=E/B=3.00\times10^6\,\mathrm{m/s}.

(b) Interpretation and further calculation

F→E=+eEŷ=3.8448×10−15ŷN.\vec F_E=+eE\hat y=3.8448\times10^{-15}\hat y\,\mathrm N.F→B=−ev0Bŷ=−3.8448×10−15ŷN.\vec F_B=-ev_0B\hat y=-3.8448\times10^{-15}\hat y\,\mathrm N.

Physical check

Both forces reverse for the opposite charge sign, so the selected speed remains E/B.

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