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Lorentz Force - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Lorentz force is considered one of the most asked concept.
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51 Questions around this concept.
Solve by difficulty
In the given fig e- enters into a magnetic field. It deflects from
A particle of charge -16 x 10-18 coulomb moving with velocity 10 ms-1 along the -axis enters a region where a magnetic field of induction B is along the
-axis, and an electric field of magnitude 104 V/m is along the negative
-axis. If the charged particle continues moving along the
-axis, the magnitude of B is
A charged particle with charge enters a region of constant, uniform, and mutually orthogonal fields
with a velocity
perpendicular to both
and comes out without any change in magnitude or direction
. Then
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In a certain region, static electric and magnetic fields exist. The magnetic field is given by $\overrightarrow{\mathrm{B}}=\mathrm{B}_0(\hat{i}+2 \hat{j}-4 \hat{k})$. If a test charge moving with a velocity $\vec{v}=v_0(3 \hat{i}-\hat{j}+2 \hat{k})$ experiences no force in that region, then the electric field in the region, in SI units, is :
An electron moving towards east enters a magnetic field directed towards the west. The force on the electron will be -
A charged particle is thrown into space with a uniform magnetic field and uniform electric field.
Choose the correct possibility of the path of the charged particle.
An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then
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) $\vec{v}, \vec{E}_{\text {and }} \vec{B}$ 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 the figure, the particle will pass through the field with the same velocity, without any deviation in the path.
$
\begin{aligned}
& q E=q v B \\
& \Rightarrow v=\frac{E}{B}
\end{aligned}
$
And in this situation, as $F_e=F_m$ i.e.
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