Lorentz Force: When Charges Dance in Magnetic Fields

Movie Hook: Thor’s Hammer in Action

Ever wonder why Thor’s hammer Mjolnir can only be lifted by the “worthy”? Imagine if it used electromagnetic forces to detect neural patterns! When charged particles (like ions in your brain) move through magnetic fields, they experience a force perpendicular to both their motion and the field. This is the Lorentz Force - and it’s responsible for everything from cosmic ray deflection to how particle accelerators work. Let’s master the dance of charges in magnetic fields!

The Big Picture

What’s the Deal?

Moving charges in magnetic fields experience a force perpendicular to velocity
This force does NO work (always perpendicular to motion)
Results in circular/helical motion - basis for cyclotrons, mass spectrometers
Combined with electric fields, gives complete electromagnetic force

JEE Perspective:

JEE Main: Circular motion, cyclotron frequency, radius calculations
JEE Advanced: Helical motion, velocity selectors, mass spectrometers, combined E-B fields

Core Concept: The Lorentz Force

Magnetic Force on Moving Charge

A charge $q$ moving with velocity $\vec{v}$ in magnetic field $\vec{B}$ experiences force:

$$\boxed{\vec{F} = q(\vec{v} \times \vec{B})}$$

Magnitude:

$$\boxed{F = qvB\sin\theta}$$

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$

Direction: Right-hand rule (for positive charge)

Fingers: velocity direction
Bend fingers: toward B direction
Thumb: force direction

For negative charge: Force is in opposite direction!

Complete Lorentz Force (Electric + Magnetic)

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

Memory Trick: “E-M Love Story”

Electric force: parallel to E (along field)
Magnetic force: perpendicular to v and B (cross product)
Together: they’re the complete electromagnetic force!

Key Properties of Magnetic Force

1. No Work Done!

Crucial Insight: Magnetic force is ALWAYS perpendicular to velocity

$$W = \vec{F} \cdot \vec{v} = q(\vec{v} \times \vec{B}) \cdot \vec{v} = 0$$

Consequences:

Kinetic energy remains constant
Only DIRECTION changes, not speed
Magnetic field cannot speed up or slow down particles!

JEE Gold: “Magnetic force changes direction, NOT speed!”

2. Force Depends on Direction

Angle θ	sin θ	Force	Motion
0° (parallel)	0	F = 0	Straight line
90° (perpendicular)	1	F = qvB (max)	Circular
Other	sin θ	F = qvB sin θ	Helical

3. Charge Sign Matters

Positive charge: Force by right-hand rule
Negative charge: Force opposite to right-hand rule
Application: Separating particles in mass spectrometers!

Standard Motion Patterns

1. Circular Motion (v ⊥ B)

When velocity is perpendicular to magnetic field:

Centripetal Force = Magnetic Force:

$$\frac{mv^2}{r} = qvB$$

Radius of circular path:

$$\boxed{r = \frac{mv}{qB}}$$

Time period:

$$\boxed{T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}}$$

Frequency (Cyclotron frequency):

$$\boxed{f = \frac{1}{T} = \frac{qB}{2\pi m}}$$

Angular frequency:

$$\boxed{\omega = 2\pi f = \frac{qB}{m}}$$

Memory Trick: “RTFO - Remember The Frequency is ωmega”

Radius: $\frac{mv}{qB}$ (has velocity)
Time period: $\frac{2\pi m}{qB}$ (no velocity!)
Frequency: $\frac{qB}{2\pi m}$ (independent of v!)
Omega: $\frac{qB}{m}$ (simplest form!)

Key Insight: Frequency is INDEPENDENT of velocity and radius!

Faster particles move in larger circles but take same time
This is the BASIS for cyclotron operation!

Interactive Demo: Lorentz Force Visualizer

Explore how charged particles move in magnetic fields with this comprehensive 3D visualizer. Adjust velocity, angle, charge sign, and magnetic field strength to see circular, helical, and straight-line motion in action!

Try these experiments:

Circular motion: Set angle to 90 degrees - particle moves in a perfect circle
Helical motion: Set angle to 45 degrees - particle spirals along B-field
Straight line: Set angle to 0 degrees - no magnetic force, no deflection
Cyclotron demo: Watch how radius increases with velocity (key to particle accelerators)

2. Helical Motion (v at angle to B)

When velocity makes angle $\theta$ with B:

Resolve velocity:

Parallel component: $v_\parallel = v\cos\theta$ (no force, constant)
Perpendicular component: $v_\perp = v\sin\theta$ (circular motion)

Radius of helix:

$$\boxed{r = \frac{mv\sin\theta}{qB}}$$

Pitch (distance traveled in one revolution):

$$\boxed{p = v_\parallel T = v\cos\theta \times \frac{2\pi m}{qB} = \frac{2\pi mv\cos\theta}{qB}}$$

Visualization:

        ↑ B (z-axis)
        |
    ----⊙---- (circular motion in x-y plane)
   /    |    \
  /     |     \
 ↗      ↑      ↗ (helical path)
       pitch

Applications:

Charged particles in Earth’s magnetic field
Particles in magnetic bottles (fusion reactors)
Aurora borealis (charged particles spiraling along field lines)

3. Straight Line Motion (v ∥ B)

When velocity is parallel or anti-parallel to B:

$\theta = 0°$ or $180°$
$\sin\theta = 0$
$F = 0$
Motion: Uniform straight line (no deflection)

Applications: Real-World Devices

1. Cyclotron (Particle Accelerator)

Principle: Charged particles spiral outward while being accelerated

Key Features:

Two D-shaped electrodes (dees)
Uniform magnetic field perpendicular to plane
Alternating electric field between dees

Frequency of AC voltage:

$$\boxed{f = \frac{qB}{2\pi m}}$$

Maximum kinetic energy:

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}$$ $$\boxed{KE_{\text{max}} = \frac{q^2B^2R^2}{2m}}$$

where R is the radius of the dees

JEE Trick: Frequency is independent of radius - that’s why it works!

2. Velocity Selector

Setup: Crossed electric and magnetic fields (E ⊥ B ⊥ v)

Condition for straight-line motion:

$$F_E = F_B$$ $$qE = qvB$$

Selected velocity:

$$\boxed{v = \frac{E}{B}}$$

Application: Only particles with this specific velocity pass through undeflected!

3. Mass Spectrometer

Components:

Velocity selector (selects specific v)
Magnetic field region (particles move in semicircle)
Detector (measures radius)

Radius in B-field:

$$r = \frac{mv}{qB}$$

Since v is selected (v = E/B):

$$r = \frac{m(E/B)}{qB} = \frac{mE}{qB^2}$$

Mass determination:

$$\boxed{m = \frac{qB^2r}{E}}$$

Application: Separating isotopes, identifying molecules!

JEE Pattern: Often asks for ratio of radii for different masses:

$$\frac{r_1}{r_2} = \frac{m_1}{m_2}$$

(for same q, E, B)

4. Magnetic Focusing

Charged particles starting from same point but different angles return to same point after half revolution

Distance between focuses:

$$\boxed{d = \frac{\pi mv}{qB}}$$

Application: CRT displays, electron microscopes

Interactive Demo: Particle Motion Simulator

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

def particle_in_B_field(q, m, v, theta, B, t_max=1e-6, dt=1e-9):
    """
    Simulate charged particle in uniform magnetic field

    Parameters:
    q: charge (C)
    m: mass (kg)
    v: speed (m/s)
    theta: angle with B-field (degrees)
    B: magnetic field strength (T)
    t_max: simulation time (s)
    dt: time step (s)
    """
    # Convert angle to radians
    theta_rad = np.radians(theta)

    # Initial velocity components
    v_perp = v * np.sin(theta_rad)  # perpendicular to B
    v_para = v * np.cos(theta_rad)  # parallel to B

    # Initialize position and velocity
    x, y, z = 0, 0, 0
    vx, vy, vz = v_perp, 0, v_para

    # Storage
    positions = [[x, y, z]]

    # Time evolution
    t = 0
    while t < t_max:
        # Lorentz force (B in z-direction)
        Fx = q * vy * B
        Fy = -q * vx * B
        Fz = 0

        # Update velocity
        vx += (Fx/m) * dt
        vy += (Fy/m) * dt
        vz += (Fz/m) * dt

        # Update position
        x += vx * dt
        y += vy * dt
        z += vz * dt

        positions.append([x, y, z])
        t += dt

    return np.array(positions)

# Example 1: Pure circular motion (theta = 90°)
q = 1.6e-19  # electron charge
m = 9.1e-31  # electron mass
v = 1e6      # 1 million m/s
B = 0.1      # Tesla

fig = plt.figure(figsize=(15, 5))

# Circular motion
ax1 = fig.add_subplot(131, projection='3d')
positions = particle_in_B_field(q, m, v, 90, B, t_max=2e-8)
ax1.plot(positions[:, 0]*1e2, positions[:, 1]*1e2, positions[:, 2]*1e2, 'b-', linewidth=2)
ax1.set_xlabel('x (cm)')
ax1.set_ylabel('y (cm)')
ax1.set_zlabel('z (cm)')
ax1.set_title('Circular Motion (θ = 90°)')

# Helical motion (theta = 45°)
ax2 = fig.add_subplot(132, projection='3d')
positions = particle_in_B_field(q, m, v, 45, B, t_max=4e-8)
ax2.plot(positions[:, 0]*1e2, positions[:, 1]*1e2, positions[:, 2]*1e2, 'r-', linewidth=2)
ax2.set_xlabel('x (cm)')
ax2.set_ylabel('y (cm)')
ax2.set_zlabel('z (cm)')
ax2.set_title('Helical Motion (θ = 45°)')

# Gentle helix (theta = 30°)
ax3 = fig.add_subplot(133, projection='3d')
positions = particle_in_B_field(q, m, v, 30, B, t_max=4e-8)
ax3.plot(positions[:, 0]*1e2, positions[:, 1]*1e2, positions[:, 2]*1e2, 'g-', linewidth=2)
ax3.set_xlabel('x (cm)')
ax3.set_ylabel('y (cm)')
ax3.set_zlabel('z (cm)')
ax3.set_title('Helical Motion (θ = 30°)')

plt.tight_layout()
plt.show()

# Calculate theoretical values
r = (m * v * np.sin(np.radians(90))) / (abs(q) * B)
T = (2 * np.pi * m) / (abs(q) * B)
f = 1/T

print(f"Radius of circular motion: {r*100:.4f} cm")
print(f"Time period: {T*1e9:.4f} ns")
print(f"Cyclotron frequency: {f*1e-6:.4f} MHz")

Experiment: Change theta to see transition from circular to helical to straight-line motion!

Common JEE Mistakes (Avoid These!)

Mistake 1: Confusing Force Direction

Wrong: Force is along velocity Right: Force is perpendicular to BOTH v and B (use right-hand rule!)

Mistake 2: Thinking Magnetic Force Does Work

Wrong: Magnetic field increases kinetic energy Right: F ⊥ v always, so W = 0, KE constant!

Mistake 3: Wrong Formula for Helical Motion

Wrong: Using r = mv/qB for helical motion Right: r = mv sin θ/qB (only perpendicular component!)

Mistake 4: Forgetting Mass Dependence

Wrong: All particles have same radius in B-field Right: r ∝ m (for same v and q)

Mistake 5: Time Period Confusion

Wrong: Time period depends on velocity Right: T = 2πm/qB (independent of v and r!)

Problem-Solving Strategy: The MOVE Method

Magnetic force direction (right-hand rule) Orient the coordinate system Velocity components (parallel and perpendicular) Equations for circular/helical motion

Practice Problems

Level 1: JEE Main Foundations

Problem 1.1: An electron (m = 9.1 × 10⁻³¹ kg, q = -1.6 × 10⁻¹⁹ C) moves with speed 10⁶ m/s perpendicular to a magnetic field of 0.1 T. Find the radius of its circular path.

Solution

Given:

m = 9.1 × 10⁻³¹ kg
v = 10⁶ m/s
B = 0.1 T
q = 1.6 × 10⁻¹⁹ C (magnitude)

Formula: $r = \frac{mv}{qB}$

$$r = \frac{9.1 \times 10^{-31} \times 10^6}{1.6 \times 10^{-19} \times 0.1}$$ $$r = \frac{9.1 \times 10^{-25}}{1.6 \times 10^{-20}} = \frac{9.1}{1.6} \times 10^{-5}$$ $$r = 5.69 \times 10^{-5} \text{ m} = 0.057 \text{ mm}$$

Answer: 0.057 mm or 57 μm

Problem 1.2: A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves in a circular path of radius 10 cm in a magnetic field of 0.5 T. Find its speed.

Solution

Given: r = 0.1 m, B = 0.5 T, m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C

From: $r = \frac{mv}{qB}$

$$v = \frac{rqB}{m} = \frac{0.1 \times 1.6 \times 10^{-19} \times 0.5}{1.67 \times 10^{-27}}$$ $$v = \frac{0.8 \times 10^{-20}}{1.67 \times 10^{-27}} = \frac{0.8}{1.67} \times 10^{7}$$ $$v = 4.79 \times 10^{6} \text{ m/s}$$

Answer: 4.79 × 10⁶ m/s (about 4.8 million m/s)

Problem 1.3: Find the cyclotron frequency for an electron in a magnetic field of 0.2 T.

Solution

Given: B = 0.2 T, m = 9.1 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C

Formula: $f = \frac{qB}{2\pi m}$

$$f = \frac{1.6 \times 10^{-19} \times 0.2}{2\pi \times 9.1 \times 10^{-31}}$$ $$f = \frac{3.2 \times 10^{-20}}{5.71 \times 10^{-30}} = 5.6 \times 10^{9} \text{ Hz}$$

Answer: 5.6 GHz (5.6 × 10⁹ Hz)

Note: Frequency is independent of velocity!

Level 2: JEE Advanced Application

Problem 2.1: A charged particle enters a region of uniform magnetic field with velocity making 30° with the field direction. If the pitch of the helical path is equal to the radius of the helix, find the speed of the particle. (Given: q/m = 10⁷ C/kg, B = 0.1 T)

Solution

Given: θ = 30°, pitch p = radius r, q/m = 10⁷ C/kg, B = 0.1 T

Radius: $r = \frac{mv\sin\theta}{qB} = \frac{v\sin 30°}{(q/m)B} = \frac{v \times 0.5}{10^7 \times 0.1} = \frac{v}{2 \times 10^6}$

Pitch: $p = \frac{2\pi mv\cos\theta}{qB} = \frac{2\pi v\cos 30°}{(q/m)B} = \frac{2\pi v \times \frac{\sqrt{3}}{2}}{10^7 \times 0.1}$

$$p = \frac{\pi\sqrt{3} v}{10^6}$$

Given p = r:

$$\frac{\pi\sqrt{3} v}{10^6} = \frac{v}{2 \times 10^6}$$ $$\pi\sqrt{3} = \frac{1}{2}$$

Wait, this gives a contradiction! Let me recalculate…

Actually: Set p = r properly:

$$\frac{2\pi v\cos 30°}{(q/m)B} = \frac{v\sin 30°}{(q/m)B}$$ $$2\pi \cos 30° = \sin 30°$$

This is not satisfied for 30°. The problem is asking for v given the condition.

Correct approach:

$$p = r$$ $$\frac{2\pi m v\cos\theta}{qB} = \frac{m v\sin\theta}{qB}$$ $$2\pi \cos\theta = \sin\theta$$ $$\tan\theta = 2\pi$$

This doesn’t match θ = 30°.

Reinterpreting: If given p = r AND θ = 30°, find v:

$$r = \frac{v \times 0.5}{10^7 \times 0.1} = \frac{v}{2 \times 10^6}$$ $$p = \frac{2\pi v \times \frac{\sqrt{3}}{2}}{10^6} = \frac{\pi\sqrt{3} v}{10^6}$$

If p = r (numerically):

$$\frac{\pi\sqrt{3} v}{10^6} = \frac{v}{2 \times 10^6}$$ $$\pi\sqrt{3} \times 2 = 1$$

This is inconsistent. The problem likely means find condition, not specific v.

Answer: The condition p = r with θ = 30° is not self-consistent. If p = r, then tan θ = 2π.

Problem 2.2: A velocity selector has E = 500 V/m and B = 0.05 T. What velocity particles will pass through undeflected? If a proton passes through, what is its kinetic energy?

Solution

Given: E = 500 V/m, B = 0.05 T

(a) Selected velocity:

$$v = \frac{E}{B} = \frac{500}{0.05} = 10,000 \text{ m/s} = 10^4 \text{ m/s}$$

(b) Kinetic energy of proton:

m = 1.67 × 10⁻²⁷ kg
v = 10⁴ m/s

$$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (10^4)^2$$ $$KE = 0.835 \times 10^{-27} \times 10^8 = 8.35 \times 10^{-20} \text{ J}$$

In eV:

$$KE = \frac{8.35 \times 10^{-20}}{1.6 \times 10^{-19}} = 0.52 \text{ eV}$$

Answers:

Selected velocity: 10⁴ m/s
Proton KE: 8.35 × 10⁻²⁰ J = 0.52 eV

Problem 2.3: Two isotopes of mass m₁ and m₂ are accelerated through the same potential V and then enter a uniform magnetic field perpendicular to their motion. Find the ratio of their radii.

Solution

Energy from potential:

$$qV = \frac{1}{2}mv^2$$ $$v = \sqrt{\frac{2qV}{m}}$$

Radius in magnetic field:

$$r = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mqV}}{qB} = \frac{\sqrt{2mV}}{B}\sqrt{\frac{q}{q}}$$ $$r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$

For same q, V, B:

$$r \propto \sqrt{m}$$

Ratio:

$$\boxed{\frac{r_1}{r_2} = \sqrt{\frac{m_1}{m_2}}}$$

Answer: $\frac{r_1}{r_2} = \sqrt{\frac{m_1}{m_2}}$

JEE Trick: For particles accelerated through same V: radius ratio = √(mass ratio)

Level 3: JEE Advanced Mastery

Problem 3.1: A particle of mass m and charge q is projected with velocity v in a region where electric field E and magnetic field B are both present and parallel. Find the trajectory of the particle.

Solution

Setup: Let both $\vec{E}$ and $\vec{B}$ be along z-axis, initial velocity $\vec{v}$ at angle θ to z-axis.

Force analysis:

Electric force: $\vec{F}_E = q\vec{E}$ (along z-axis)
Magnetic force: $\vec{F}_B = q\vec{v} \times \vec{B}$

Key insight: If $\vec{v} \parallel \vec{B}$, then $\vec{F}_B = 0$

Components:

$v_z$ (parallel to B): Experiences only electric force
$v_\perp$ (perpendicular to B): Experiences magnetic force (circular motion)

Motion in z-direction:

$$F_z = qE$$ $$z = \frac{1}{2}\frac{qE}{m}t^2 + v_z t$$

(uniformly accelerated)

Motion in x-y plane: Circular with radius $r = \frac{mv_\perp}{qB}$ and frequency $\omega = \frac{qB}{m}$

Combined trajectory: Helix with increasing pitch (accelerating helix)

Parametric equations:

$$x = r\cos(\omega t)$$ $$y = r\sin(\omega t)$$ $$z = v_z t + \frac{1}{2}\frac{qE}{m}t^2$$

Answer: Accelerating helical motion (helix with increasing pitch)

Problem 3.2: A cyclotron has dees of radius 50 cm and magnetic field 1.5 T. (a) What is the maximum energy of accelerated protons? (b) If the frequency of the applied voltage is 20 MHz, what particle is being accelerated?

Solution

(a) Maximum energy:

At maximum radius R = 0.5 m:

$$v_{\max} = \frac{qBR}{m} = \frac{1.6 \times 10^{-19} \times 1.5 \times 0.5}{1.67 \times 10^{-27}}$$ $$v_{\max} = \frac{1.2 \times 10^{-19}}{1.67 \times 10^{-27}} = 7.19 \times 10^{7} \text{ m/s}$$ $$KE_{\max} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (7.19 \times 10^7)^2$$ $$KE_{\max} = 4.32 \times 10^{-12} \text{ J}$$

In MeV:

$$KE = \frac{4.32 \times 10^{-12}}{1.6 \times 10^{-19} \times 10^6} = 27 \text{ MeV}$$

(b) Particle identification:

Cyclotron frequency: $f = \frac{qB}{2\pi m}$

Given f = 20 MHz = 20 × 10⁶ Hz:

$$20 \times 10^6 = \frac{q \times 1.5}{2\pi m}$$ $$m = \frac{1.5q}{2\pi \times 20 \times 10^6} = \frac{1.5q}{1.257 \times 10^8}$$

For proton (q = 1.6 × 10⁻¹⁹ C):

$$m = \frac{1.5 \times 1.6 \times 10^{-19}}{1.257 \times 10^8} = 1.91 \times 10^{-27} \text{ kg}$$

This is close to proton mass (1.67 × 10⁻²⁷ kg), but let’s check deuteron:

For deuteron (m ≈ 2 × 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C):

$$f = \frac{1.6 \times 10^{-19} \times 1.5}{2\pi \times 3.34 \times 10^{-27}} = 11.4 \text{ MHz}$$

Not matching. It’s approximately a proton.

Answers:

(a) Maximum KE = 27 MeV
(b) Proton (given the approximations)

Problem 3.3: A particle with specific charge q/m enters a region of crossed electric and magnetic fields (E perpendicular to B). Initially at rest, describe its motion. The fields are: E = 100 V/m (x-direction), B = 0.5 T (z-direction).

Solution

Initial condition: v = 0 at t = 0

Forces:

Electric: $F_E = qE$ (constant, x-direction)
Magnetic: $F_B = q\vec{v} \times \vec{B}$ (perpendicular to v)

Motion analysis:

Phase 1: Initially, only electric force acts (v = 0, so F_B = 0)

Particle accelerates in x-direction
Gains velocity: $v_x = \frac{qE}{m}t$

Phase 2: Once moving, magnetic force appears

$F_{B,y} = qv_xB$ (in y-direction)
Particle curves in x-y plane

Equations of motion:

$$m\frac{dv_x}{dt} = qE - qv_yB$$ $$m\frac{dv_y}{dt} = qv_xB$$

This is a cycloidal motion!

Solution characteristics:

Particle drifts in y-direction with average velocity: $v_{\text{drift}} = \frac{E}{B}$
Superimposed circular motion with cyclotron frequency: $\omega = \frac{qB}{m}$

Trajectory: Cycloid (like a rolling wheel)

Drift velocity:

$$\boxed{v_{\text{drift}} = \frac{E}{B} = \frac{100}{0.5} = 200 \text{ m/s}}$$

Answer: Cycloidal motion with drift velocity 200 m/s in y-direction

Connections: The Web of Knowledge

Link to Circular Motion

Magnetic force provides centripetal force
Same equations: F = mv²/r
See: Circular Motion

Link to Current-Carrying Conductor

Current = moving charges, experiences force F = BIL
Same underlying principle
See: Force on Current Conductor

Link to Electromagnetic Induction

Moving charges in B-field is basis for motional EMF
v × B appears in both contexts
See: Motional EMF

Link to Electric Field

Complete Lorentz force: F = q(E + v × B)
E does work, B doesn’t
See: Electric Field

Quick Revision: Formula Sheet

Quantity	Formula	Note
Magnetic force	$F = qvB\sin\theta$	Max when v ⊥ B
Radius (circular)	$r = \frac{mv}{qB}$	Perpendicular motion
Time period	$T = \frac{2\pi m}{qB}$	Independent of v!
Frequency	$f = \frac{qB}{2\pi m}$	Cyclotron frequency
Radius (helical)	$r = \frac{mv\sin\theta}{qB}$	Perpendicular component
Pitch	$p = \frac{2\pi mv\cos\theta}{qB}$	Parallel component
Velocity selector	$v = \frac{E}{B}$	For straight motion
Cyclotron KE	$KE = \frac{q^2B^2R^2}{2m}$	At maximum radius

Remember: Magnetic force does NO work! (Always perpendicular to v)

Exam Strategy

Time Management:

Direction questions: 1 minute (right-hand rule)
Radius/period calculations: 2-3 minutes
Cyclotron/spectrometer: 5-7 minutes

Quick Checks:

Is force perpendicular to v? (Always yes!)
Does speed change? (No, only direction!)
Is frequency independent of v? (Yes!)

Common Patterns:

40% circular motion problems
30% cyclotron/mass spectrometer
20% helical motion
10% crossed fields

Summary: Key Takeaways

Lorentz Force: $\vec{F} = q(\vec{v} \times \vec{B})$ - always perpendicular to velocity
No Work Done: Magnetic force changes direction, NOT speed (KE constant)
Circular Motion: Radius $r = \frac{mv}{qB}$, frequency $f = \frac{qB}{2\pi m}$ (independent of v!)
Helical Motion: When v makes angle θ with B, combines circular + linear motion
Applications: Cyclotron (accelerator), mass spectrometer (isotope separation), velocity selector
JEE Gold: Frequency is independent of velocity - this is the KEY to cyclotron!

Last Updated: March 22, 2025 Next Topic: Force on Current-Carrying Conductor Previous Topic: Ampere’s Circuital Law

Lorentz Force: When Charges Dance in Magnetic Fields

Movie Hook: Thor’s Hammer in Action

The Big Picture

Core Concept: The Lorentz Force

Magnetic Force on Moving Charge

Complete Lorentz Force (Electric + Magnetic)

Key Properties of Magnetic Force

1. No Work Done!

2. Force Depends on Direction

3. Charge Sign Matters

Standard Motion Patterns

1. Circular Motion (v ⊥ B)

Interactive Demo: Lorentz Force Visualizer

2. Helical Motion (v at angle to B)

3. Straight Line Motion (v ∥ B)

Applications: Real-World Devices

1. Cyclotron (Particle Accelerator)

2. Velocity Selector

3. Mass Spectrometer

4. Magnetic Focusing

Interactive Demo: Particle Motion Simulator

Common JEE Mistakes (Avoid These!)

Mistake 1: Confusing Force Direction

Mistake 2: Thinking Magnetic Force Does Work

Mistake 3: Wrong Formula for Helical Motion

Mistake 4: Forgetting Mass Dependence

Mistake 5: Time Period Confusion

Problem-Solving Strategy: The MOVE Method

Practice Problems

Level 1: JEE Main Foundations

Level 2: JEE Advanced Application

Level 3: JEE Advanced Mastery

Connections: The Web of Knowledge

Link to Circular Motion

Link to Current-Carrying Conductor

Link to Electromagnetic Induction

Link to Electric Field

Quick Revision: Formula Sheet

Exam Strategy

Summary: Key Takeaways

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