Physics Electromagnetism - How to Solve: Magnetic Force on Moving Charge & Current (Lorentz Force, Cyclotron, Torque on Loop) – IIT JEE Guide

How to Solve: Magnetic Force on Moving Charge & Current (Lorentz Force, Cyclotron, Torque on Loop) – IIT JEE Guide

Introduction

Mastering magnetic force on charges and currents unlocks 5-7 marks in IIT JEE (Main + Advanced) every year—enough to push you from a 90th to a 99th percentile rank. It’s the foundation for cyclotrons, motors, and particle accelerators, and it appears in both numericals and theory-based questions.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand: 1. Vector cross product (right-hand rule, direction of F = q(v × B)). 2. Circular motion (centripetal force, radius, angular frequency). 3. Current as moving charges (I = q/t, drift velocity).

If any of these are shaky, stop now and review them first.

KEY TERMS & FORMULAS

1. Lorentz Force (Force on a Moving Charge)

Formula: F = q(v × B) - F = Magnetic force (N) - q = Charge (C) - v = Velocity of charge (m/s) - B = Magnetic field (T) - × = Cross product (direction given by right-hand rule)

MEMORISE THIS: - If v ⊥ B, force is maximum (F = qvB). - If v ∥ B, force is zero. - Direction: Right-hand rule (thumb = F, index = v, middle = B for positive charge).

2. Cyclotron Frequency (Angular Frequency of Charged Particle in Magnetic Field)

Formula: ω = qB/m - ω = Angular frequency (rad/s) - q = Charge (C) - B = Magnetic field (T) - m = Mass of particle (kg)

MEMORISE THIS: - Time period T = 2π/ω = 2πm/qB (independent of velocity!). - Radius of path: r = mv/qB (for v ⊥ B).

3. Force on a Current-Carrying Wire

Formula: F = I(L × B) - F = Force (N) - I = Current (A) - L = Length vector of wire (m, direction = current flow) - B = Magnetic field (T)

MEMORISE THIS: - If L ⊥ B, F = ILB. - If L ∥ B, F = 0. - Direction: Right-hand rule (thumb = F, index = I, middle = B).

4. Torque on a Current Loop (Magnetic Dipole Moment)

Formula: τ = NIAB sinθ - τ = Torque (N·m) - N = Number of turns - I = Current (A) - A = Area of loop (m²) - B = Magnetic field (T) - θ = Angle between A (normal to loop) and B

MEMORISE THIS: - Magnetic dipole moment (μ) = NIA (direction = right-hand rule around loop). - τ = μ × B (vector form). - Maximum torque when θ = 90° (loop plane ∥ B). - Zero torque when θ = 0° (loop plane ⊥ B).

STEP-BY-STEP METHOD

Step 1: Identify the Scenario

• Is it a moving charge? → Use F = q(v × B).
• Is it a current-carrying wire? → Use F = I(L × B).
• Is it a current loop in a magnetic field? → Use τ = NIAB sinθ.

Step 2: Determine Directions (Right-Hand Rule)

• For positive charges/wires:
• Index finger = Velocity (v) or Current (I).
• Middle finger = Magnetic field (B).
• Thumb = Force (F).
• For negative charges: Reverse the direction of F.

Step 3: Calculate Magnitude

• If v ⊥ B or L ⊥ B, use F = qvB or F = ILB.
• If θ is given, use F = qvB sinθ or F = ILB sinθ.
• For torque, use τ = NIAB sinθ.

Step 4: Solve for Unknowns

• Cyclotron problems: Use r = mv/qB or ω = qB/m.
• Torque problems: Find μ = NIA, then τ = μB sinθ.

Step 5: Check Units & Consistency

• Force (N) = kg·m/s².
• B (T) = N/(A·m).
• Torque (N·m) = kg·m²/s².

WORKED EXAMPLES

Example 1 – Basic (Force on a Moving Charge)

Question: An electron (q = -1.6 × 10⁻¹⁹ C) moves with velocity v = 3 × 10⁶ m/s perpendicular to a magnetic field B = 0.5 T. Find the magnitude and direction of the force.

Solution: 1. Identify scenario: Moving charge → F = q(v × B). 2. Magnitude: Since v ⊥ B, F = qvB.
- F = (1.6 × 10⁻¹⁹ C)(3 × 10⁶ m/s)(0.5 T) = 2.4 × 10⁻¹³ N. 3. Direction:
- Electron is negative, so reverse right-hand rule.
- v (right), B (into page) → F (downward).

Answer: 2.4 × 10⁻¹³ N, downward.

What we did and why: - Used F = qvB because v ⊥ B. - Reversed direction for negative charge. - Confirmed units (N = C·m/s·T).

Example 2 – Medium (Cyclotron Frequency & Radius)

Question: A proton (m = 1.67 × 10⁻²⁷ kg, q = +1.6 × 10⁻¹⁹ C) enters a B = 2 T field at v = 4 × 10⁶ m/s perpendicular to B. Find: (a) Radius of circular path. (b) Time period of revolution.

Solution: 1. Radius (r = mv/qB):
- r = (1.67 × 10⁻²⁷ kg)(4 × 10⁶ m/s) / (1.6 × 10⁻¹⁹ C)(2 T) = 0.0209 m = 2.09 cm. 2. Time period (T = 2πm/qB):
- T = 2π(1.67 × 10⁻²⁷) / (1.6 × 10⁻¹⁹)(2) = 3.28 × 10⁻⁸ s.

Answer: (a) 2.09 cm (b) 3.28 × 10⁻⁸ s

What we did and why: - Used r = mv/qB for circular motion. - Used T = 2πm/qB (independent of velocity). - Checked units (m, s).

Example 3 – Exam-Style (Torque on a Current Loop)

Question: A rectangular loop of 10 turns, carrying 5 A, has dimensions 10 cm × 20 cm. It is placed in a B = 0.2 T field such that the normal to the loop makes 30° with B. Find the torque.

Solution: 1. Area (A) = 0.1 m × 0.2 m = 0.02 m². 2. Magnetic dipole moment (μ) = NIA = 10 × 5 A × 0.02 m² = 1 A·m². 3. Torque (τ = μB sinθ) = (1)(0.2 T)(sin 30°) = 0.1 N·m.

Answer: 0.1 N·m

What we did and why: - Calculated μ = NIA first. - Used τ = μB sinθ (not cosθ!). - Confirmed θ = angle between μ and B.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for magnetic force:

1. Moving charge? F = q(v × B). Right-hand rule for direction. If charge is negative, flip the force.
2. Current in a wire? F = I(L × B). Same right-hand rule.
3. Cyclotron? r = mv/qB, T = 2πm/qB—velocity doesn’t matter!
4. Torque on a loop? τ = NIAB sinθ. θ is between the normal and B, not the loop plane.
5. Always check units—Tesla (T) = N/(A·m). If your answer is in cm, convert to meters.
6. Examiners love angles—if they give θ, use sinθ or cosθ correctly.

Now go solve 3 problems—right-hand rule, cyclotron, and torque—and you’ll own this topic."