An electron (mass 9×10⁻³¹ kg and charge 1.6×10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field ⃗B of magnitude 9×10⁻⁴ T perpendicular to its direction of motion. We wish to apply a uniform electric field ⃗E together with the magnetic field so that the electron d…

Question

An electron (mass 9×10⁻³¹ kg and charge 1.6×10⁻¹⁹ C) moving with speed c/100 (c = speed of light) is injected into a magnetic field ⃗B of magnitude 9×10⁻⁴ T perpendicular to its direction of motion. We wish to apply a uniform electric field ⃗E together with the magnetic field so that the electron does not deflect from its path. Then

Accepted Answer

(4) ⃗E is perpendicular to ⃗B and its magnitude is 27×10² V m⁻¹. Given: v=c/(100)=(3×10⁸)/(100)=3×10⁶ m s⁻¹; B=9×10⁻⁴ T, perpendicular to ⃗v. Idea: the magnetic force q⃗v×⃗B is perpendicular to both ⃗v and ⃗B, so the electric force must point exactly opposite to it. That fixes the direction: ⃗E must lie along ⃗v×⃗B, which is perpendicular to ⃗B (and to ⃗v). Balancing the magnitudes: qE=qvB, so E=vB. E=3×10⁶×9×10⁻⁴. E=2.7×10³=27×10² V m⁻¹.

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