📝 Problem Types & Solved Examples
Every problem pattern you'll face in CBSE, NEET, JEE Main & Advanced
Step 1: Identify what examiner is testing
Step 2: Select concept/formula
Step 3: Check direction/sign
Step 4: Solve systematically
Step 5: Verify units & reasonableness
Type 1: Direct Formula Application
What examiner tests: Can you identify and apply correct formula?
CBSE/NEET Favorite
Question: A proton (q = 1.6 × 10-19 C) moves with velocity 106 m/s perpendicular to a magnetic field of 0.5 T. Calculate the magnetic force on it.
Given:
- q = 1.6 × 10-19 C
- v = 106 m/s
- B = 0.5 T
- θ = 90° (perpendicular)
What Examiner Tests:
Can you identify this as Lorentz force problem and apply F = qvB sin θ?
Concept Selection:
Solution:
F = (1.6 × 10-19) × (106) × (0.5) × sin(90°)
F = 1.6 × 10-19 × 106 × 0.5 × 1
F = 8 × 10-14 N
Direction:
Use Fleming's Left Hand Rule: Perpendicular to both v and B
For perpendicular: sin 90° = 1, so F = qvB directly
Question: A long straight wire carries current of 10 A. Find magnetic field at distance 5 cm from the wire.
Given:
- I = 10 A
- r = 5 cm = 0.05 m
- μ₀ = 4π × 10-7 T·m/A
Concept Selection:
Solution:
B = (4π × 10-7 × 10) / (2π × 0.05)
B = (4 × 10-6) / (0.1)
B = 4 × 10-5 T = 40 μT
They give I and r in different units (like mA and mm). Convert to SI first!
Question: An electron (m = 9.1 × 10-31 kg, q = 1.6 × 10-19 C) moving at 107 m/s enters perpendicular to magnetic field 0.01 T. Find radius of circular path.
Given:
- m = 9.1 × 10-31 kg
- q = 1.6 × 10-19 C
- v = 107 m/s
- B = 0.01 T
Concept Selection:
Solution:
r = (9.1 × 10-31 × 107) / (1.6 × 10-19 × 0.01)
r = (9.1 × 10-24) / (1.6 × 10-21)
r = 5.69 × 10-3 m ≈ 5.7 mm
Lighter particles (electrons) have smaller radius than heavier particles (protons) at same v and B.
Type 2: Conceptual Problems
What examiner tests: Do you understand WHY, not just formula?
JEE Main/NEET Favorite
Question: Explain why magnetic force on a moving charged particle does no work. What remains constant?
Concept:
Magnetic force F⃗ = q(v⃗ × B⃗) is always perpendicular to velocity v⃗
Reasoning:
Work = F⃗ · ds⃗ = F⃗ · (v⃗ dt)
Since F⃗ ⊥ v⃗, their dot product = 0
Therefore, W = 0
Consequence:
- Kinetic energy remains constant
- Speed remains constant
- Only direction changes (circular/helical motion)
They say "force is perpendicular" but don't explain WHY that means zero work. Always connect: perpendicular → dot product zero → W = 0
Question: Derive and explain why time period of charged particle in magnetic field is independent of its velocity.
Derivation:
Step 1: Radius r = mv/(qB)
Step 2: Time period T = 2πr/v
Step 3: Substitute r:
T = 2π[mv/(qB)]/v = 2πm/(qB)
Result: T is independent of v!
Physical Meaning:
• Faster particle → larger radius
• But also covers larger distance in same time
• These effects exactly cancel out
This is WHY cyclotron works. Particle can be accelerated repeatedly at same frequency regardless of speed.
Question: Two parallel wires carry equal currents. In which case do they attract: (a) Same direction (b) Opposite direction? Explain.
Analysis:
Case 1: Same direction currents
• Wire 1 creates field B₁ around it
• At wire 2 location, B₁ is in specific direction
• Current I₂ experiences force F₂ = B₁I₂L
• Direction: Toward wire 1 (attractive)
Case 2: Opposite direction currents
• By same analysis, force is away from each other
• Result: Repulsive
Conclusion:
Parallel currents attract, antiparallel currents repel
Students memorize "same → attract" without understanding. In exam, if asked WHY, they can't explain. Always know the mechanism!
Type 3: Multi-Step Problems
What examiner tests: Can you break complex problem into steps?
JEE Main/Advanced Level
Question: A particle (q = +e, m) enters region with E = 10³ N/C (downward) and B = 0.5 T (into page) with horizontal velocity. Find velocity for straight-line motion. If v is doubled, find radius of resulting circular path.
Part 1: Velocity for straight-line
Condition: Electric and magnetic forces balance
qE = qvB
v = E/B = 10³/0.5 = 2000 m/s
Part 2: Radius when v is doubled
New velocity: v' = 2v = 4000 m/s
Now magnetic force > electric force
Net force causes circular motion
But wait! Particle doesn't move in pure circle because electric field is still present.
This is cycloidal motion, not circular. The question tests if you know pure circular motion requires ONLY magnetic field.
Question: A galvanometer has resistance 50 Ω and gives full-scale deflection for 1 mA. Calculate shunt resistance to convert it to ammeter reading up to 1 A. What fraction of total current passes through galvanometer?
Given:
- R = 50 Ω
- Ig = 1 mA = 10-3 A
- I = 1 A (desired range)
Step 1: Find shunt S
S = (10-3 × 50) / (1 - 10-3)
S = 0.05 / 0.999 ≈ 0.050 Ω
Step 2: Fraction through galvanometer
Fraction = Ig/I = 10-3/1 = 0.001 = 0.1%
This exact question (5 marks) appears every 2-3 years. Also practice voltmeter conversion!
Type 4: Graph-Based Problems
What examiner tests: Can you extract info from graphs and apply concepts?
JEE Main Favorite
- B vs r for straight wire (hyperbolic: B ∝ 1/r)
- B vs I (linear: B ∝ I)
- r vs v for circular motion (linear: r ∝ v)
- r vs √K for charged particle (linear: r ∝ √K)
Problem: Analyzing B vs r graph
Question: Graph shows B vs 1/r for a straight wire. Slope is 2 × 10-7 T·m. Find current in wire.
Solution:
For straight wire: B = (μ₀I)/(2πr) = (μ₀I/2π) × (1/r)
This is linear in 1/r with slope = μ₀I/(2π)
Given slope = 2 × 10-7
μ₀I/(2π) = 2 × 10-7
I = 2 × 10-7 × 2π / μ₀
I = 2 × 10-7 × 2π / (4π × 10-7)
I = 1 A
Type 5: Assertion-Reason
What examiner tests: Do you know BOTH fact AND reasoning?
NEET/CBSE Favorite
Options:
(A) Both assertion and reason true, reason is correct explanation
(B) Both true, reason not correct explanation
(C) Assertion true, reason false
(D) Assertion false, reason true
Assertion: Magnetic force cannot change speed of charged particle.
Reason: Magnetic force is always perpendicular to velocity.
Analysis:
Assertion: TRUE (F ⊥ v → W = 0 → ΔKE = 0 → speed constant)
Reason: TRUE (F = q(v × B) → F ⊥ v always)
Connection: Reason EXPLAINS assertion (perpendicular → no work → constant speed)
Answer: (A)
Assertion: Magnetic field inside ideal solenoid is uniform.
Reason: All field lines inside solenoid are parallel and equally spaced.
Analysis:
Assertion: TRUE (B = μ₀nI, constant everywhere inside)
Reason: TRUE (parallel and equally spaced lines → uniform field)
Connection: Reason EXPLAINS assertion
Answer: (A)
Type 6: Case-Based (Paragraph Type)
What examiner tests: Can you extract info from passage and apply?
CBSE Term-2 Pattern
Case Study: Cyclotron
A cyclotron is a device used to accelerate charged particles to high energies. It consists of two D-shaped hollow metal chambers called dees, placed in uniform magnetic field perpendicular to their plane. High-frequency alternating voltage is applied between the dees. Charged particle moves in semicircular path inside each dee and gets accelerated each time it crosses the gap.
Given: B = 1.5 T, radius of dee R = 0.5 m, particle = proton (m = 1.67 × 10-27 kg, q = 1.6 × 10-19 C)
Q1: Calculate cyclotron frequency
Solution:
f = qB/(2πm)
f = (1.6 × 10-19 × 1.5)/(2π × 1.67 × 10-27)
f ≈ 2.3 × 107 Hz = 23 MHz
Q2: Maximum kinetic energy
Solution:
KEmax = q²B²R²/(2m)
KEmax = (1.6 × 10-19)² × (1.5)² × (0.5)² / (2 × 1.67 × 10-27)
KEmax ≈ 1.1 × 10-11 J ≈ 6.9 MeV
Q3: Why doesn't cyclotron work for electrons at high speed?
Answer: At relativistic speeds, mass increases (m → γm). Frequency f = qB/(2πm) decreases. Particle goes out of sync with alternating voltage.