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Lorentz Force - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Lorentz force is considered one of the most asked concept.

  • 46 Questions around this concept.

Solve by difficulty

A particle of charge -16 x 10-18 coulomb moving with velocity 10 ms-1 along the x-axis enters a region where a magnetic field of induction B is along the y-axis, and an electric field of magnitude 104 V/m is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is

A charged particle with charge q enters a region of constant, uniform, and mutually orthogonal fields \vec{E}\; and \; \vec{B} with a velocity \vec{v} perpendicular to both \vec{E}\; and \; \vec{B} and comes out without any change in magnitude or direction \vec{v}. Then

 In a certain region static electric and magnetic fields exist.  The magnetic field is given by

 If a test charge moving with a velocity experiences no force in that region, then the electric field in the region, in SI units, is :

 

In a certain region of space, electric field \vec{E} and magnetic field \vec{B} are perpendicular to each other and an electron enters a region perpendicular to the direction of \vec{B} and \vec{E} both and moves undeflected, then the velocity of the electron is

Concepts Covered - 1

Lorentz force

Lorentz force

When the moving charged particle is subjected simultaneously to both electric field \vec{E}  and magnetic field \vec{B}, the moving charged
particle will experience electric \vec{F}_{e}=q E  and magnetic force \vec{F}_{m}=q(\vec{v} \times \vec{B});  so the net force on it will be
\vec{F}=q[\vec{E}+(\vec{v} \times \vec{B})]  Which is our lorentz-force equation. Depending on the directions of \vec{v}, E  and \vec{B}  following situations are possible.

(i) When \vec{v},  \vec{E} and \vec{B}  all the three are collinear : In this situation the magnetic force on it will be zero and only electric force will act
and so  \ \vec{a}=\frac{\vec{F}}{m}=\frac{q \vec{E}}{m}.

(ii) The particle will pass through the field following a straight-line path (parallel field) with change in its speed. So in this situation speed, velocity, momentum and kinetic energy all will change without change in direction of motion as shown


(iii) \vec{v}, \vec{E} and \vec{B}  are mutually perpendicular : in this situation if \vec{E} and \vec{B}  are such that \vec{F}=\vec{F}_{e}+\vec{F}_{m}=0  ie.
\vec{a}=(\vec{F} / m)=0

as shown in figure, the particle will pass through the field with same velocity, without any deviation in path.
And in this situation, as F_{e}=F_{m}  i.e.  q E=q v B v=\frac{E}{B}.

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Lorentz force

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