Magnetism Example Problems- Solutions Included
Why Magnetism Problems Trip You Up
Most students see a magnetism problem and freeze. They stare at the diagram, scribble random formulas, and hope something sticks. It doesn't work that way.
Magnetism isn't hard. It has rules. Learn the rules, apply them systematically, and these problems become routine. This guide cuts through the confusion with real example problems and step-by-step solutions.
The Formulas You Actually Need
Before touching any problem, memorize these. Not "kind of know." Memorize them.
- Magnetic force on a moving charge: F = qvB sinθ
- Magnetic force on a current-carrying wire: F = ILB sinθ
- Magnetic field from a long straight wire: B = (μ₀I)/(2πr)
- Magnetic field inside a solenoid: B = μ₀nI
- Magnetic flux: Φ = BA cosθ
- Force between two parallel wires: F/L = (μ₀I₁I₂)/(2πd)
The symbols: q = charge, v = velocity, B = magnetic field, I = current, L = wire length, r/d = distance, n = turns per unit length, A = area, μ₀ = 4π × 10⁻⁷ T·m/A.
Problem 1: Force on a Moving Charge
Question: A proton moves at 3.0 × 10⁶ m/s perpendicular to a magnetic field of 0.5 T. What force acts on the proton?
Step 1: Identify the formula. We have a moving charge in a magnetic field → F = qvB sinθ
Step 2: Note the angle. "Perpendicular" means θ = 90°, so sinθ = 1.
Step 3: Plug in values.
q = 1.6 × 10⁻¹⁹ C (proton charge)
v = 3.0 × 10⁶ m/s
B = 0.5 T
F = (1.6 × 10⁻¹⁹)(3.0 × 10⁶)(0.5)(1)
F = 2.4 × 10⁻¹³ N
Direction? Use the right-hand rule. Point fingers in velocity direction (thumb), curl toward B (palm). Positive charge → push from palm. Proton deflects perpendicular to both v and B.
Problem 2: Electron in a Magnetic Field
Question: An electron travels at 2.0 × 10⁵ m/s parallel to a 0.4 T magnetic field. Calculate the magnetic force.
Step 1: Check the angle. Parallel means θ = 0° or 180°.
Step 2: sin(0°) = 0.
F = qvB sinθ = (1.6 × 10⁻¹⁹)(2.0 × 10⁵)(0.4)(0) = 0 N
No force. When charge moves parallel to magnetic field lines, nothing happens. The field doesn't "grab" particles moving along its lines.
Problem 3: Wire in a Magnetic Field
Question: A 0.5 m wire carrying 10 A sits perpendicular to a 0.8 T magnetic field. Find the force.
Step 1: Formula for current-carrying wire: F = ILB sinθ
Step 2: Perpendicular → sinθ = 1
F = (10)(0.5)(0.8)(1) = 4 N
Direction? Use right-hand rule again, but now thumb points in current direction, fingers in B direction. Force comes out of your palm.
Problem 4: Magnetic Field from a Wire
Question: A long straight wire carries 25 A. What is the magnetic field strength 5 cm away?
Step 1: Use B = (μ₀I)/(2πr)
Step 2: Convert distance: 5 cm = 0.05 m
B = (4π × 10⁻⁷ × 25) / (2π × 0.05)
B = (1.0 × 10⁻⁴) / (0.1) = 1.0 × 10⁻⁴ T = 0.1 mT
Direction? Wrap your right hand around the wire with thumb pointing in current direction. Fingers curl in the direction of B. Field circles the wire.
Problem 5: Solenoid Magnetic Field
Question: A solenoid has 500 turns in a 0.2 m length and carries 3 A current. Find the field strength inside.
Step 1: Find n (turns per meter): n = 500/0.2 = 2500 turns/m
Step 2: Apply B = μ₀nI
B = (4π × 10⁻⁷)(2500)(3)
B = 9.4 × 10⁻³ T = 9.4 mT
Problem 6: Force Between Two Parallel Wires
Question: Two wires 0.1 m apart each carry 20 A in the same direction. What force per meter acts on each wire?
Step 1: Formula: F/L = (μ₀I₁I₂)/(2πd)
Step 2: Plug in
F/L = (4π × 10⁻⁷ × 20 × 20) / (2π × 0.1)
F/L = (1.6 × 10⁻⁴) / (0.2) = 8 × 10⁻⁴ N/m
Attraction or repulsion? Currents in the same direction → attraction. Opposite directions → repulsion. This is how circuit breakers detect overloads.
Problem 7: Magnetic Flux
Question: A square loop (side 0.1 m) sits in a 2 T field perpendicular to the loop. Calculate the flux.
Step 1: Area = (0.1)² = 0.01 m²
Step 2: Perpendicular → θ = 0°, cosθ = 1
Φ = BA cosθ = (2)(0.01)(1) = 0.02 Wb
Quick Reference: Field Direction Rules
| Scenario | Right-Hand Rule | Force Direction |
|---|---|---|
| Moving positive charge | Thumb = v, fingers = B | Out of palm |
| Moving negative charge | Thumb = v, fingers = B | Into palm (opposite) |
| Current in wire | Thumb = current, fingers curl = B | Perpendicular to palm |
| Solenoid | Curl fingers = current, thumb = N pole | Field lines through center |
Common Mistakes That Cost You Points
- Forgetting sinθ. Always check the angle between v and B (or I and B). Parallel = zero force. Perpendicular = maximum force.
- Wrong sign for electrons. q = -1.6 × 10⁻¹⁹ C. The negative sign matters for direction.
- Confusing B and F. B is field strength. F is force experienced. Different formulas, different units.
- Using wrong formula. Charge motion → F = qvB. Wire current → F = ILB. Wire producing field → B = μ₀I/2πr. Know which situation you're in.
- Skipping unit conversion. Centimeters to meters. Millitesla to tesla. Sloppy units = wrong answer.
Getting Started: Your Problem-Solving Checklist
Run through this every time:
- Read the problem twice. Identify what's given (q, v, B, I, L, r, etc.) and what's asked (F, B, Φ, etc.)
- Pick the right formula. Don't force a formula. Match it to the situation.
- Check the angle. Perpendicular? Parallel? Something else? This determines sinθ.
- Plug in numbers with units. Convert everything to base SI units first.
- Calculate. Show your work. Partial credit exists for a reason.
- State direction. Force problems without direction are incomplete.
One More Thing
If a problem says "calculate the radius of the path," you need this: r = mv/qB. Derive it from centripetal force = magnetic force. That's a common follow-up question.
These seven problems cover the core scenarios you'll see. Practice until the right-hand rule becomes automatic. That's the skill that makes or breaks your score on this unit.