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Lorentz force is considered one of the most asked concept.
21 Questions around this concept.
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
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 :
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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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