The force on a charged particle created by its motion in a magnetic field is maximum at what angle between the particle velocity and field?

Question

BlackTom AI · Accepted Answer

Consider the magnetic force on a moving charge, F = q v × B. The magnitude is F = q v B sin(theta), where theta is the angle between the velocity vector v and the magnetic field B.
Option 0o: If the velocity is parallel to the field (theta = 0°), sin(0°) = 0, so the magnetic force is zero. This makes sense because there is no component of velocity perpendicular to the field to experience a magnetic deflection.
Option 180o: If the velocity is anti-parallel to the field (theta = 180°), sin(180°) = 0 as well, so again the magnetic force is zero for the same reason as in 0°, since there is no perpendicular component to interact with the field.
Option 90o: When theta = 90°, the velocity is perpendicular to the magnetic field. In this case sin(90°) = 1, giving the maximum possible magnitude for the magnetic force (F = q v B). This corresponds to the largest transverse deflection before other effects or constraints come into play.
Option 45o: At theta = 45°, sin(45°) = √2/2, so the force magnitude is less than the maximum and smaller than at 90°. While this represents a valid nonzero magnetic force, it is not the maximum possible value.
In summary, the magnetic force magnitude depends on sin(theta); it reaches its peak when the velocity is perpendicular to the magnetic field, and becomes zero when the velocity is aligned with the field in either direction. The other angles yield nonzero but not maximal forces.

The force on a charged particle created by its motion in a magnetic field is maximum at what angle between the particle velocity and field?Single choice

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