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✍️ Practice Section

Test your understanding with timed practice sets for every difficulty level

Practice Timer

30:00
Select Difficulty:

Easy Level - CBSE Boards 15 Questions | 30 min

Q1. A proton moving with velocity 10⁶ m/s enters perpendicular to magnetic field 0.2 T. Force on proton? ▼

Options:

(A) 1.6 × 10⁻¹⁴ N
(B) 3.2 × 10⁻¹⁴ N
(C) 6.4 × 10⁻¹⁴ N
(D) 8.0 × 10⁻¹⁴ N

Answer: (B) 3.2 × 10⁻¹⁴ N

Solution:

F = qvB sin θ

F = (1.6 × 10⁻¹⁹) × (10⁶) × (0.2) × sin(90°)

F = 1.6 × 10⁻¹⁹ × 10⁶ × 0.2 = 3.2 × 10⁻¹⁴ N

Q2. Magnetic field at distance 2 cm from long wire carrying 5 A current? ▼

Options:

(A) 2 × 10⁻⁵ T
(B) 4 × 10⁻⁵ T
(C) 5 × 10⁻⁵ T
(D) 10 × 10⁻⁵ T

Answer: (C) 5 × 10⁻⁵ T

Solution:

B = μ₀I/(2πr)

B = (4π × 10⁻⁷ × 5)/(2π × 0.02)

B = (2 × 10⁻⁶)/0.04 = 5 × 10⁻⁵ T

Q3. Time period of electron (m = 9.1 × 10⁻³¹ kg) in magnetic field 0.01 T? ▼

Options:

(A) 3.57 × 10⁻⁹ s
(B) 5.68 × 10⁻⁹ s
(C) 7.14 × 10⁻⁹ s
(D) 9.28 × 10⁻⁹ s

Answer: (A) 3.57 × 10⁻⁹ s

Solution:

T = 2πm/(qB)

T = (2π × 9.1 × 10⁻³¹)/(1.6 × 10⁻¹⁹ × 0.01)

T ≈ 3.57 × 10⁻⁹ s

🎯
Complete Practice Set

12 more questions available. Focus on:

  • Direct formula application
  • Unit conversion
  • Standard numericals

Moderate Level - JEE Main/NEET 20 Questions | 45 min

Q1. Two particles with q/m ratios 1:2 enter same B field with same velocity. Ratio of radii? ▼

Options:

(A) 1:2
(B) 2:1
(C) 1:1
(D) 1:4

Answer: (B) 2:1

Solution:

r = mv/(qB) = (m/q) × (v/B)

Since v and B are same:

r ∝ m/q

r₁/r₂ = (m/q)₁/(m/q)₂ = (q/m)₂/(q/m)₁ = 2/1

Q2. Particle enters B field at 30° angle with velocity 10⁶ m/s. Find pitch of helix (m = 10⁻²⁶ kg, q = 10⁻¹⁹ C, B = 0.1 T) ▼

Options:

(A) 1.09 × 10⁻⁴ m
(B) 5.44 × 10⁻⁴ m
(C) 1.09 × 10⁻³ m
(D) 5.44 × 10⁻³ m

Answer: (B) 5.44 × 10⁻⁴ m

Solution:

Pitch = v cos θ × T

T = 2πm/(qB) = (2π × 10⁻²⁶)/(10⁻¹⁹ × 0.1) = 6.28 × 10⁻⁷ s

Pitch = 10⁶ × cos(30°) × 6.28 × 10⁻⁷

Pitch = 10⁶ × 0.866 × 6.28 × 10⁻⁷ ≈ 5.44 × 10⁻⁴ m

Q3. Galvanometer (R = 100 Ω, Ig = 1 mA) to be converted to voltmeter reading 10 V. Required resistance? ▼

Options:

(A) 9900 Ω
(B) 10000 Ω
(C) 10100 Ω
(D) 10900 Ω

Answer: (A) 9900 Ω

Solution:

For voltmeter, high resistance in series

Total resistance = V/Ig = 10/10⁻³ = 10000 Ω

Series resistance = Total - R = 10000 - 100 = 9900 Ω

Advanced Level - JEE Advanced 10 Questions | 60 min

Q1. Charged particle in combined E (downward, 10³ N/C) and B (into page, 0.5 T). If v = 3000 m/s horizontal, find trajectory curvature. ▼

This requires:

  • Finding net force
  • Understanding cycloidal motion
  • Vector analysis

Hint:

• First check if v = E/B (straight line condition)

• Here v = 3000 m/s but E/B = 2000 m/s

• So particle will curve. Magnetic force dominates.

• Net acceleration has both components

Q2. Minimum B field to confine particle (KE = 100 eV, q = e, m = mp) within 1 m radius? ▼

Approach:

r = mv/(qB) ≤ R

Need to find v from KE, then find Bmin

Solution:

KE = 100 eV = 100 × 1.6 × 10⁻¹⁹ J = 1.6 × 10⁻¹⁷ J

½mpv² = 1.6 × 10⁻¹⁷

v = √(2 × 1.6 × 10⁻¹⁷/1.67 × 10⁻²⁷) ≈ 1.38 × 10⁵ m/s

Bmin = mpv/(eR)

Bmin = (1.67 × 10⁻²⁷ × 1.38 × 10⁵)/(1.6 × 10⁻¹⁹ × 1)

Bmin ≈ 1.44 × 10⁻³ T = 1.44 mT

🧠
Advanced Practice Strategy

Don't just solve. After each question, ask:

  • Can I solve it in alternate way?
  • What if parameters were different?
  • How does answer change in limiting cases?

Full Chapter Mock Test 25 Questions | 60 min

Question Distribution:

  • Easy: 10 questions (2 marks each)
  • Moderate: 10 questions (3 marks each)
  • Hard: 5 questions (5 marks each)

Total: 75 marks

🎯
Mock Test Strategy

Time allocation:

  • Easy (20 min): Quick answers, move fast
  • Moderate (25 min): Careful calculation
  • Hard (15 min): Attempt best 3 out of 5, skip rest

Target: 60/75 (80%) for JEE Advanced readiness

← Advanced Thinking Next: Exam Strategy →