Lorentz Force

The Lorentz Force is a fundamental concept in electromagnetism that describes the force experienced by a charged particle moving through an electric and magnetic field. This force, named after the Dutch physicist Hendrik Lorentz, plays a crucial role in understanding the behavior of charged particles in various physical contexts. Mathematically, it is expressed as F = q(E + v × B), where F is the force, q is the charge, E is the electric field, v is the velocity, and B is the magnetic field. In real life, the Lorentz Force is observed in numerous applications, such as in the operation of electric motors and generators, where it helps convert electrical energy into mechanical energy and vice versa. It's also fundamental to the working of devices like cyclotrons and mass spectrometers, which are used in medical treatments and scientific research. Even in our daily lives, the principles behind the Lorentz Force can be seen in the functioning of devices like smartphones and televisions, making it an essential concept in both advanced technology and everyday gadgets. In this article, we will discuss the concept of Lorentz force and solve examples for concept clarity.

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1. Lorentz Force
2. Solved Examples Based on Lorentz Force
3. Summary

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 $\vec{E}$ 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 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)$\vec{v}, \vec{E}$ 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 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. $\begin{aligned} & q E=q v B \\ & \Rightarrow v=\frac{E}{B}\end{aligned}$.

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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

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

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

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

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 -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

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

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

The field of induction B is along -the axis.

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

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

Resolve the motion along the three-coordinate axis

$\begin{aligned} & \therefore \quad a_x=\frac{F_x}{m}=\frac{q}{m}\left(E_x+v_y B_z-v_z B_y\right) \\ & a_y=\frac{F_y}{m}=\frac{q}{m}\left(E_y+v_z B_x-v_x B_z\right) \\ & a_z=\frac{F_z}{m}=\frac{q}{m}\left(E_z+v_x B_y-v_y B_x\right) \\ & \text { Since } E_x=E_y=0, v_y=v_z=0, \quad B_x=B_z=0 \\ & \therefore \quad a_x=a_y=0, \quad a_z=\frac{q}{m}\left(-E_z+v_x B_y\right)\end{aligned}$

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

$\begin{aligned} & \therefore \quad E_z=v_x B_y \\ & { }_{\text {or }} B_y=\frac{E_z}{v_x}=\frac{10^4}{10}=10^3 \frac{\mathrm{Wb}}{\mathrm{m}^2}\end{aligned}$

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{aligned}
& \quad q E=q v B \\
& \text { i.e. } \Rightarrow v=\frac{E}{B} .
\end{aligned}
$

In putting values we get

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

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.