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Worked Examples · Example 4.2

Q.If the magnetic field is parallel to the positive yy-axis and the charged particle is moving along the positive xx-axis (Fig. 4.4), which way would the Lorentz force be for

(a) an electron (negative charge),
(b) a proton (positive charge).
Figure 4.4
Figure 4.4
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The Lorentz force direction is given by F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}). For a proton (positive qq), the force is along +z+z; for an electron (negative qq), the force is along −z-z.

The core idea here is the direction of the Lorentz force, set by the cross product v⃗×B⃗\vec{v}\times\vec{B} together with the sign of the charge. The Lorentz force law tells us that a charged particle moving in a magnetic field experiences a force perpendicular to both its velocity and the field. This perpendicular nature is what makes magnetic forces so elegant (and tricky): they never speed up or slow down a particle, only bend its path.

For this problem, we have:

  • B⃗\vec{B} along +y+y (positive y-axis)
  • v⃗\vec{v} along +x+x (positive x-axis)
  • Two cases: q=+eq = +e (proton) and q=−eq = -e (electron)

The right-hand rule (or the cross product) gives the direction of v⃗×B⃗\vec{v} \times \vec{B}, and then the sign of qq decides whether the force is along that direction or opposite to it.


  1. Find the direction of v⃗×B⃗\vec{v} \times \vec{B}

    Using the right-hand rule: point fingers along v⃗\vec{v} (positive xx), curl them toward B⃗\vec{B} (positive yy). Your thumb points along +z+z (upward).

    Mathematically: x^×y^=z^\hat{x} \times \hat{y} = \hat{z}, so v⃗×B⃗\vec{v} \times \vec{B} points along +z+z.

  2. Apply the Lorentz force law

    The force is F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}).

    • For a proton (q=+eq = +e): F⃗=+e(along +z)\vec{F} = +e(\text{along }+z), so the force is along +z+z.
    • For an electron (q=−eq = -e): F⃗=−e(along +z)\vec{F} = -e(\text{along }+z), so the force is along −z-z. …

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