Magnetic Forces & Charged Particle Trajectories: Master Notes

Chapter Overview

Magnetic forces deflect moving charges without altering their kinetic energy. In IIT-JEE (Advanced) and Olympiads, prime questions test helical trajectory pitch and radius, cycloidal paths in crossed $\overrightarrow{E}\times \overrightarrow{B}$ fields, effective length vectors (${\overrightarrow{L}}_{\text{eff}}$) on arbitrary wires, and magnetic mirror confinement.

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1. The Lorentz Force & Trajectory Classifications

$$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$$

Fundamental Principle: The magnetic force ${\overrightarrow{F}}_B=q(\overrightarrow{v}\times \overrightarrow{B})$ is always perpendicular to velocity $\overrightarrow{v}$. Therefore:

$${\overrightarrow{F}}_B\cdot \overrightarrow{v}=0⟹P=0⟹\text{Magnetic force does ZERO work on a free charge!}$$

(Kinetic energy and speed remain strictly constant under pure magnetic fields).

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1.1 Trajectory Regimes in Uniform $\overrightarrow{B}$

Multi-Mode DiagramMagnetic Forces: Lorentz Helical & Cycloidal Trajectories

Option 1: Publication-Grade Scientific Vector SVG

Charged particle trajectory in magnetic field B: Radius R = mv / (qB) and helical pitch p = 2pi * m * v_parallel / (qB).

Cyclotron Radius: $r = \frac{m v_\perp}{q B}$

Helical Pitch: $p = v_\parallel T = \frac{2\pi m v_\parallel}{q B}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate 3D helical particle spiral in magnetic field.

Magnetic Field B: 4 T

Radius ($r$):2.50 cm

Option 3: Interactive 3D Spatial Viewport

3D helical trajectory viewport.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Lorentz Force:$\vec{F} = q(\vec{v}\times\vec{B})$

Option 5: Reactive Parametric Physics with Live Calculus

Cyclotron frequency & radius calculator.

Magnetic Field B: 5 T

: 45

Frequency ($f_c$):$76.0 MHz$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

Initial Velocity Condition	Resulting Trajectory	Key Formulas
$\overrightarrow{v}\parallel \overrightarrow{B}$ ($\theta =0^{\circ },{180}^{\circ }$)	Straight Line	Unaltered velocity $\overrightarrow{v}(t)={\overrightarrow{v}}_0$
$\overrightarrow{v}\perp \overrightarrow{B}$ ($\theta ={90}^{\circ }$)	Circle in plane $\perp \overrightarrow{B}$	Radius $r_c=\frac{mv}{qB}=\frac{p}{qB}=\frac{\sqrt{2mK}}{qB}$, Period $T=\frac{2\pi m}{qB}$
$\overrightarrow{v}$ at angle $\theta$ to $\overrightarrow{B}$	Helix	Radius $r=\frac{mv\sin \theta }{qB}$, Pitch $p=v_{\parallel }T=\frac{2\pi mv\cos \theta }{qB}$

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2. Motion in Crossed $\overrightarrow{E}$ and $\overrightarrow{B}$ Fields ($\overrightarrow{E}\perp \overrightarrow{B}$)

2.1 Velocity Selector

When $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular, particles pass undeflected if:

$$qE=qvB⟹\boxed{{\displaystyle v=\frac{E}{B}}}$$

2.2 Cycloidal Motion (Released from Rest at Origin in $\overrightarrow{E}=E\hat{j}$, $\overrightarrow{B}=B\hat{k}$)

$$x(t)=\frac{E}{\omega B}(\omega t-\sin \omega t),y(t)=\frac{E}{\omega B}(1-\cos \omega t)$$

• Maximum height attained along $y$: $y_{max}=\frac{2E}{\omega B}=\frac{2mE}{q{B}^2}$
• Maximum speed: $v_{max}=\frac{2E}{B}$

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3. Magnetic Force on Current-Carrying Wires

$$\overrightarrow{F}=\int I(d\overrightarrow{\ell }\times \overrightarrow{B})=I({\overrightarrow{L}}_{\text{eff}}\times \overrightarrow{B})$$

Where ${\overrightarrow{L}}_{\text{eff}}=\int d\overrightarrow{\ell }$ is the straight-line displacement vector connecting the start point to the end point!

<!-- Endpoints --> <circle cx="60" cy="90" r="5" fill="var(--vp-c-brand-1)"/> <text x="45" y="115" fill="var(--vp-c-brand-1)" font-size="13" font-weight="700">A</text> <circle cx="360" cy="90" r="5" fill="var(--vp-c-brand-1)"/> <text x="370" y="115" fill="var(--vp-c-brand-1)" font-size="13" font-weight="700">B</text> <!-- Effective Vector L_eff --> <line x1="60" y1="90" x2="350" y2="90" stroke="#3b82f6" stroke-width="2.5" stroke-dasharray="4 3" marker-end="url(#arrow-blue)"/> <text x="180" y="115" fill="#3b82f6" font-size="13" font-weight="700">L_eff Vector</text> 

Figure Effective wire length vector L_eff: Magnetic force on any arbitrary curved wire in uniform B depends solely on the straight-line displacement between endpoints A and B.

• Closed Loop in Uniform $\overrightarrow{B}$: Since ${\overrightarrow{L}}_{\text{eff}}=\oint d\overrightarrow{\ell }=\overrightarrow{0}⟹\boxed{{\displaystyle {\overrightarrow{F}}_{\text{net}}=\overrightarrow{0}}}$.

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 (Paper 1) — Cycloid Height in Crossed Fields

Question:
A particle of mass $m$ and charge $+q$ is released from rest at the origin $(0,0,0)$ in a region with uniform electric field $\overrightarrow{E}=E_0\hat{j}$ and magnetic field $\overrightarrow{B}=B_0\hat{k}$. Find the maximum $y$-coordinate $y_{max}$ reached by the particle and its speed at that point.

Step-by-Step Solution:

1. Work-Energy Theorem: Since only the electric field does work:

   $$W_{\text{net}}=q{E}_{0}y=\frac{1}{2}m{v}^2⟹v(y)=\sqrt{\frac{2q{E}_{0}y}{m}}\text{--- (1)}$$

2. At Maximum Height $y_{max}$: The vertical velocity component becomes zero ($v_y=0$). Therefore, the total velocity is purely horizontal: $\overrightarrow{v}=v_x\hat{i}$. In the cycloidal trajectory frame, the drift velocity is $v_d=\frac{E_0}{B_0}$, and at the apex the speed is twice the drift speed:

   $$v_{max}=2v_d=\frac{2E_0}{B_0}$$

3. Equating with Work-Energy (1):

   $$\frac{1}{2}m{\left(\frac{2E_0}{B_0}\right)}^2=q{E}_0{y}_{max}$$$$\frac{1}{2}m\frac{4E_0^2}{B_0^2}=q{E}_0{y}_{max}⟹\boxed{{\displaystyle y_{max}=\frac{2m{E}_0}{q{B}_0^2}}}$$$$\boxed{{\displaystyle v(y_{max})=\frac{2E_0}{B_0}}}$$

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PYQ 2: JEE Advanced 2019 — Force on Semicircular Wire

Question:
A rigid wire carrying current $I$ consists of a semicircle of radius $R$ in the $xy$-plane lying between $(-R,0)$ and $(+R,0)$. A uniform magnetic field $\overrightarrow{B}=B_0\hat{k}$ is applied. Find the net magnetic force on the wire.

Step-by-Step Solution:

Using the effective length vector concept:

$${\overrightarrow{L}}_{\text{eff}}={\overrightarrow{r}}_{\text{final}}-{\overrightarrow{r}}_{\text{initial}}=(R\hat{i})-(-R\hat{i})=2R\hat{i}$$$$\overrightarrow{F}=I({\overrightarrow{L}}_{\text{eff}}\times \overrightarrow{B})=I(2R\hat{i}\times B_0\hat{k})=2IR{B}_0(\hat{i}\times \hat{k})=\boxed{{\displaystyle -2IR{B}_0\hat{j}}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Cyclotron Radius	$r_c=\frac{m{v}_{\perp }}{qB}=\frac{\sqrt{2mK}}{qB}$	Perpendicular motion
Helix Pitch	$p=\frac{2\pi m{v}_{\parallel }}{qB}$	Distance along $\overrightarrow{B}$ in 1 turn
Force on Arbitrary Wire	$\overrightarrow{F}=I({\overrightarrow{L}}_{\text{eff}}\times \overrightarrow{B})$	End-to-end vector
Force Between Wires	$\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$	Parallel currents attract
Magnetic Work	$W_B=0$	Speed is always invariant