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Magnetism And Matter Class 12 Physics Revision — Grandmaster Guide

A

Ayush (Founder)

Exam Strategist

Last Updated: 2026-06-01
  • B = μ₀nI
  • F = IL × B
  • F = qv × B
  • τ = nIAB
  • μ = μ₀ × (n × I × A)
  • M = nIA
  • B = μ₀H + μ₀M
  • χ = (μ - 1)
  • μ = (1 + χ)
  • B = μ₀(H + M)
  • Magnetic field on axis of loop = μ₀I/(2r)
  • Magnetic field on perpendicular bisector of loop = μ₀I/(2πr³)³
  • Magnetic field due to solenoid = μ₀nI
  • Magnetic field due to toroid = μ₀nI/(2πr)
  • Force per unit length = μ₀I₁I₂/(2πr)

🪤 The 5 Mistakes That Cost Marks

  • Not using the correct formula for magnetic field due to a current carrying wire
  • Forgetting to consider the direction of magnetic field while calculating force on a current carrying wire
  • Not understanding the concept of magnetic moment and its relation to magnetic field
  • Incorrectly applying the formula for magnetic field due to a solenoid
  • Not considering the effect of magnetic field on a moving charge

✏️ 3 Solved PYQs

  • A current of 10 A is flowing ∈ a wire of length 1 m. The wire is placed ∈ a magnetic field of 0.1 T. If the angle between the wire and magnetic field is 30°, calculate the force acting on the wire.
    • F = IL × B = 10 × 1 × 0.1 × s∈(30°) = 0.5 N
  • A solenoid of length 10 cm and radius 1 cm has 1000 turns. If a current of 5 A is flowing through it, calculate the magnetic field at its center.
    • B = μ₀nI = 4π × 10⁻⁷ × 1000 × 5 = 0.0628 T
  • A bar magnet of magnetic moment 10 A m² is placed ∈ a magnetic field of 0.1 T. If the angle between the magnet and magnetic field is 60°, calculate the torque acting on the magnet.
    • τ = MB s∈(θ) = 10 × 0.1 × s∈(60°) = 0.866 Nm

🧠 The One Thing Most Students Get Wrong

  • The concept of magnetic moment and its relation to magnetic field. Many students get confused between the two and end up using the wrong formula.

👁️ Ayush's Note

  • To solve problems related to magnetism, always start by identifying the given quantities and the unknown quantity.
  • Use the formula bank to find the relevant formula for the problem.
  • Always consider the direction of magnetic field and the angle between the wire and magnetic field while calculating force.
  • Practice problems regularly to develop a strong understanding of the concepts.

🔁 Last 5 Minutes Box

  • Revise the formula for magnetic field due to a current carrying wire.
  • Go through the concept of magnetic moment and its relation to magnetic field.
  • Practice a few problems related to force on a current carrying wire and magnetic field due to a solenoid.
  • Make sure to understand the concept of torque and its relation to magnetic moment.

📝 Practice MCQs

1. What is the magnetic field at the center of a solenoid of length 10 cm and radius 1 cm with 1000 turns, if a current of 5 A is flowing through it?

A) 0.01 T

B) 0.0628 T

C) 0.1 T

D) 1 T

Answer: B) 0.0628 T. Explanation: B = μ₀nI = 4π × 10⁻⁷ × 1000 × 5 = 0.0628 T

2. A current of 10 A is flowing ∈ a wire of length 1 m. The wire is placed ∈ a magnetic field of 0.1 T. If the angle between the wire and magnetic field is 30°, what is the force acting on the wire?

A) 0.1 N

B) 0.5 N

C) 1 N

D) 5 N

Answer: B) 0.5 N. Explanation: F = IL × B = 10 × 1 × 0.1 × s∈(30°) = 0.5 N

3. A bar magnet of magnetic moment 10 A m² is placed ∈ a magnetic field of 0.1 T. If the angle between the magnet and magnetic field is 60°, what is the torque acting on the magnet?

A) 0.5 Nm

B) 0.866 Nm

C) 1 Nm

D) 5 Nm

Answer: B) 0.866 Nm. Explanation: τ = MB s∈(θ) = 10 × 0.1 × s∈(60°) = 0.866 Nm

4. What is the magnetic field due to a current carrying loop of radius 1 m and current 5 A at a point on its axis at a distance of 2 m from its center?

A) 0.01 T

B) 0.0628 T

C) 0.1 T

D) 1 T

Answer: A) 0.01 T. Explanation: B = μ₀I/(2r) = 4π × 10⁻⁷ × 5/(2 × 2) = 0.01 T

5. A toroid of radius 10 cm and 1000 turns is carrying a current of 5 A. What is the magnetic field at its center?

A) 0.01 T

B) 0.0628 T

C) 0.1 T

D) 1 T

Answer: B) 0.0628 T. Explanation: B = μ₀nI/(2πr) = 4π × 10⁻⁷ × 1000 × 5/(2π × 0.1) = 0.0628 T


🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

🎬 Watch video explanations on YouTube →


This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.


📚 Related Topics

Continue your revision with these related guides:

  • 📖 Atoms Class 12 Physics Revision — Grandmaster Guide
  • 📖 Communication Systems Class 12 Physics Revision — Grandmaster Guide
  • 📖 Dual Nature Of Radiation Class 12 Physics Revision — Grandmaster Guide
  • 📖 Electromagnetic Waves Class 12 Physics Revision — Grandmaster Guide
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Prepared for Scholar
Date: 2026-06-01
CATEGORY: Exam Notes
  • B = μ₀nI
  • F = IL × B
  • F = qv × B
  • τ = nIAB
  • μ = μ₀ × (n × I × A)
  • M = nIA
  • B = μ₀H + μ₀M
  • χ = (μ - 1)
  • μ = (1 + χ)
  • B = μ₀(H + M)
  • Magnetic field on axis of loop = μ₀I/(2r)
  • Magnetic field on perpendicular bisector of loop = μ₀I/(2πr³)³
  • Magnetic field due to solenoid = μ₀nI
  • Magnetic field due to toroid = μ₀nI/(2πr)
  • Force per unit length = μ₀I₁I₂/(2πr)

🪤 The 5 Mistakes That Cost Marks

  • Not using the correct formula for magnetic field due to a current carrying wire
  • Forgetting to consider the direction of magnetic field while calculating force on a current carrying wire
  • Not understanding the concept of magnetic moment and its relation to magnetic field
  • Incorrectly applying the formula for magnetic field due to a solenoid
  • Not considering the effect of magnetic field on a moving charge

✏️ 3 Solved PYQs

  • A current of 10 A is flowing ∈ a wire of length 1 m. The wire is placed ∈ a magnetic field of 0.1 T. If the angle between the wire and magnetic field is 30°, calculate the force acting on the wire.
    • F = IL × B = 10 × 1 × 0.1 × s∈(30°) = 0.5 N
  • A solenoid of length 10 cm and radius 1 cm has 1000 turns. If a current of 5 A is flowing through it, calculate the magnetic field at its center.
    • B = μ₀nI = 4π × 10⁻⁷ × 1000 × 5 = 0.0628 T
  • A bar magnet of magnetic moment 10 A m² is placed ∈ a magnetic field of 0.1 T. If the angle between the magnet and magnetic field is 60°, calculate the torque acting on the magnet.
    • τ = MB s∈(θ) = 10 × 0.1 × s∈(60°) = 0.866 Nm

🧠 The One Thing Most Students Get Wrong

  • The concept of magnetic moment and its relation to magnetic field. Many students get confused between the two and end up using the wrong formula.

👁️ Ayush's Note

  • To solve problems related to magnetism, always start by identifying the given quantities and the unknown quantity.
  • Use the formula bank to find the relevant formula for the problem.
  • Always consider the direction of magnetic field and the angle between the wire and magnetic field while calculating force.
  • Practice problems regularly to develop a strong understanding of the concepts.

🔁 Last 5 Minutes Box

  • Revise the formula for magnetic field due to a current carrying wire.
  • Go through the concept of magnetic moment and its relation to magnetic field.
  • Practice a few problems related to force on a current carrying wire and magnetic field due to a solenoid.
  • Make sure to understand the concept of torque and its relation to magnetic moment.

📝 Practice MCQs

1. What is the magnetic field at the center of a solenoid of length 10 cm and radius 1 cm with 1000 turns, if a current of 5 A is flowing through it?

A) 0.01 T

B) 0.0628 T

C) 0.1 T

D) 1 T

Answer: B) 0.0628 T. Explanation: B = μ₀nI = 4π × 10⁻⁷ × 1000 × 5 = 0.0628 T

2. A current of 10 A is flowing ∈ a wire of length 1 m. The wire is placed ∈ a magnetic field of 0.1 T. If the angle between the wire and magnetic field is 30°, what is the force acting on the wire?

A) 0.1 N

B) 0.5 N

C) 1 N

D) 5 N

Answer: B) 0.5 N. Explanation: F = IL × B = 10 × 1 × 0.1 × s∈(30°) = 0.5 N

3. A bar magnet of magnetic moment 10 A m² is placed ∈ a magnetic field of 0.1 T. If the angle between the magnet and magnetic field is 60°, what is the torque acting on the magnet?

A) 0.5 Nm

B) 0.866 Nm

C) 1 Nm

D) 5 Nm

Answer: B) 0.866 Nm. Explanation: τ = MB s∈(θ) = 10 × 0.1 × s∈(60°) = 0.866 Nm

4. What is the magnetic field due to a current carrying loop of radius 1 m and current 5 A at a point on its axis at a distance of 2 m from its center?

A) 0.01 T

B) 0.0628 T

C) 0.1 T

D) 1 T

Answer: A) 0.01 T. Explanation: B = μ₀I/(2r) = 4π × 10⁻⁷ × 5/(2 × 2) = 0.01 T

5. A toroid of radius 10 cm and 1000 turns is carrying a current of 5 A. What is the magnetic field at its center?

A) 0.01 T

B) 0.0628 T

C) 0.1 T

D) 1 T

Answer: B) 0.0628 T. Explanation: B = μ₀nI/(2πr) = 4π × 10⁻⁷ × 1000 × 5/(2π × 0.1) = 0.0628 T


🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

🎬 Watch video explanations on YouTube →


This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.


📚 Related Topics

Continue your revision with these related guides:

  • 📖 Atoms Class 12 Physics Revision — Grandmaster Guide
  • 📖 Communication Systems Class 12 Physics Revision — Grandmaster Guide
  • 📖 Dual Nature Of Radiation Class 12 Physics Revision — Grandmaster Guide
  • 📖 Electromagnetic Waves Class 12 Physics Revision — Grandmaster Guide