Lorentz Force

Lorentz Force

The Lorentz Force is a cornerstone concept in electromagnetism, describing the combined electric and magnetic force on a charged particle. Named after the Dutch physicist Hendrik Lorentz, it is mathematically expressed as F = q(E + v × B), where F represents the force, q the charge of the particle, E the electric field, v the velocity of the particle, and B the magnetic field. This principle is vital for understanding how charged particles move in different fields and has significant practical implications.`

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

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

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

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

as shown in the figure, the particle will pass through the field with the same velocity, without any deviation in the path.
In this situation, as $F_e=F_m$ i.e. $\begin{array}{rl}& qE=qvB\\ & \Rightarrow v=\frac{E}{B}\end{array}$.

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Solved Examples Based on Lorentz Force

Example 1: A charged particle is moving with velocity v in a magnetic field of induction B. The force on the particle will be maximum when

1) $v$ and $B$ are in the same direction.
2) v and B are in opposite directions.
3) v and $B$ are perpendicular
4) $v$ and $B$ are at an angle of ${45}^{\circ }$

Solution:

Lorentz Force

A particle of charge q moving with velocity V in the presence of an electric field E and a magnetic field B experiences a force

Lorentz force in magnetic field

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$ or $|F|=qvB\sin \theta$

F will be the maximum when v and B are perpendicular.

Hence, the answer is the option (2).

Example 2: In the given fig $e^{-}$ enters into a magnetic field. It deflects from

1)+ve X-direction.

2)-ve X-direction.

3)+ve Y-direction.

4) -ve Y-direction.

Solution:

Lorentz Force

$\underset{F}{\overrightarrow{v}}=q[\underset{E}{\overrightarrow{V}}+(\overrightarrow{V}\times \underset{B}{\overrightarrow{A}})]$

wherein

Lorentz equation

From Fleming's left-hand rule, the electron deflects in a negative y-direction.

Hence, the answer is the option (4).

Example 3: 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

1)10³ Wb/m²

2)10⁵ Wb/m²

3)10¹⁶ Wb/m²

4)10^-3 Wb/m²

Solution:

Magnetic field If V(vector), E (vector), and B (vector) are mutually perpendicular

$\begin{array}{rl}& {\mathrm{F}}_{\mathrm{e}}={\mathrm{F}}_{\mathrm{m}}\\ & v=\frac{E}{B}\end{array}$

A particle travels along $x$-the axis. Hence $v_y=v_z=0$.

The field of induction B is along $y$-the axis.

The electric field is along the negative z-axis. $E_x=E_y=0$

$\therefore$ Net force on particle $\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

Resolve the motion along the three-coordinate axis

$\begin{array}{rl}& \therefore a_x=\frac{F_x}{m}=\frac{q}{m}(E_x+v_y{B}_z-v_z{B}_y)\\ & a_y=\frac{F_y}{m}=\frac{q}{m}(E_y+v_z{B}_x-v_x{B}_z)\\ & a_z=\frac{F_z}{m}=\frac{q}{m}(E_z+v_x{B}_y-v_y{B}_x)\\ & \text{Since}{E}_x=E_y=0,v_y=v_z=0,B_x=B_z=0\\ & \therefore a_x=a_y=0,a_z=\frac{q}{m}(-E_z+v_x{B}_y)\end{array}$

Again $a_z=0$ as the particle traverses through the region undeflected.

$\begin{array}{rl}& \therefore E_z=v_x{B}_y\\ & {}_{\text{or}}{B}_y=\frac{E_z}{v_x}=\frac{{10}^4}{10}={10}^3\frac{\mathrm{Wb}}{{\mathrm{m}}^2}\end{array}$

Hence the magnitude of B is 10³ wb/m²

Hence, the answer is the option (4).

Example 4: A beam of $\alpha$ particles passes through mutually perpendicular electric and magnetic fields undeflected with the speed $1.5\times {10}^8\mathrm{cm}/\mathrm{s}$. If the electric field is $90\mathrm{kV}{\mathrm{m}}^{-1}$, the magnetic field in the region is

1) $6\times {10}^{-4}\mathrm{T}$
2) $6\times {10}^{-2}\mathrm{T}$
3) $6\times {10}^{-3}\mathrm{T}$
4) $6\times {10}^{-6}T$

Solution:

The particle will pass through the field with the same velocity, without any deviation in the path.

In this situation, as $F_e=F_m$

$\begin{array}{rl}& qE=qvB\\ & \text{i.e.}\Rightarrow v=\frac{E}{B}.\end{array}$

In putting values we get

$\begin{array}{rl}v& =\frac{E}{B}\\ B& =\frac{E}{v}\\ B& =\frac{90\times {10}^3}{1.5\times {10}^8}\\ B& =6\times {10}^{-2}T\end{array}$

Hence, the answer is the option (2).

Summary

The Lorentz Force is a key concept in electromagnetism, describing the force experienced by a charged particle moving through electric and magnetic fields. it has practical implications in various fields, from the operation of electric motors and generators to devices like cyclotrons and mass spectrometers. Depending on the relative directions of v, E, and B, the particle's motion can vary, demonstrating the versatility and significance of this fundamental force in both advanced technology and everyday applications.