Which of the following is true about the Lorentz force?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → It depends on the velocity of the moving particle.
The Lorentz force on a charge \( q \) moving with velocity \( \vec{v} \) in an electric field \( \vec{E} \) and a magnetic field \( \vec{B} \) is \( \vec{F}=q\,(\vec{E}+\vec{v}\times\vec{B}) \).
Option (1) It is the same in magnitude for an alpha particle and an electron moving with the same velocity. — Incorrect. The force is proportional to the charge. An alpha particle carries \( 2e \) and an electron \( e \) in magnitude. So at the same velocity, the force on the alpha particle is twice as large.
Option (2) It does not depend upon the nature of the electric or magnetic field. — Incorrect. Both parts of the force depend on the fields. The part \( q\vec{E} \) depends on the electric field. The part \( q\,\vec{v}\times\vec{B} \) depends on the size and direction of the magnetic field.
Option (3) It depends on the velocity of the moving particle. — Correct. The magnetic part \( q\,\vec{v}\times\vec{B} \) depends on the speed and direction of motion of the particle. It is zero when \( \vec{v} \) is parallel to \( \vec{B} \). It is largest when \( \vec{v} \) is perpendicular to \( \vec{B} \).
Option (4) For a charged particle at rest, the Lorentz force is always zero. — Incorrect. At rest, only the magnetic part becomes zero. In an electric field, the particle still feels the force \( q\vec{E} \).
NCERT (2026-27):
Magnetic Field, Lorentz Force – “… The force on an electric charge q due to both of them can be written as
\( \mathbf{F}=q\,[\,\mathbf{E}(\mathbf{r})+\mathbf{v}\times\mathbf{B}(\mathbf{r})\,]\equiv\mathbf{F}_{\mathrm{electric}}+\mathbf{F}_{\mathrm{magnetic}} \qquad (4.3) \)
This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. …
(i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge.”