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If a conductor is charged from an exterior electric field, moved in a magnetic field perpendicular to the E-Field, the velocity's direction is also perpendicular to both the B-field and E-field.

  • How would the Lorentz force disrupt the accumulated charges by the edges?
  • Could the Lorentz force surpass the electrostatic forces?
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How would the Lorentz force disrupt the accumulated charges by the edges?

It should decrease or increase charge separation depending on the direction of the magnetic field (up or down).

Could the Lorentz force surpass the electrostatic forces?

It can't "surpass" the electrostatic forces, because electrostatic forces are responsible for the separation of charges that brings about the Lorentz force in the first place.

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  • $\begingroup$ The first statement makes perfect sense, but I'm confused with the following: "because electrostatic forces are responsible for the separation of charges that brings about the Lorentz force in the first place." How does the seperated charge bring about the Lorentz force? Isn't it F = qE + qv x B $\endgroup$ – UnkownConstants Aug 3 '18 at 22:22
  • $\begingroup$ To my understanding, the motion of the conductor with the charges within it, bring about the Lorentz force. $\endgroup$ – UnkownConstants Aug 3 '18 at 22:24
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    $\begingroup$ @UnkownConstants Without the electric field, the conductor would be neutral and there won't be any charge q the Lorentz force would act on. When the electric field is applied, electrons are pushed to one side of the conductor leaving positive ions on the other: now we have two moving charges the Lorentz force could act on. $\endgroup$ – V.F. Aug 3 '18 at 22:31
  • $\begingroup$ „It should decrease or increase charge separation depending on the direction of the magnetic field (up or down).“ Shouldn’t the external field deflect the charges - electrons in on side and protons in the other and rotate by this a little bit the body? $\endgroup$ – HolgerFiedler Aug 4 '18 at 7:47
  • $\begingroup$ @HolgerFiedler The magnetic field should exert equal forces on both positive and negative charges, so the net force on the conductor should be zero. $\endgroup$ – V.F. Aug 4 '18 at 13:59

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