Magnetic Field and Torque?
A bar magnet of length 10 cm is placed inside a solenoid of 100 turns per meter. If the solenoid carries a current of 2 A, what is the magnitude of the torque experienced by the bar magnet? Assume the magnetic moment of the bar magnet is 0.5 A·m².
1 Answer
📌 CONCEPT: The magnetic field inside a solenoid is given by the formula B = μ₀nI, where μ₀ is the magnetic constant, n is the number of turns per unit length, and I is the current flowing through the solenoid. The torque experienced by a bar magnet in a magnetic field is given by the formula τ = μB sinθ, where μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between the magnetic moment and the magnetic field.
📐 RULE / FORMULA: The magnetic field inside a solenoid is given by B = μ₀nI and the torque experienced by a bar magnet is given by τ = μB sinθ.
💡 WORKED EXAMPLE: Given μ = 0.5 A·m², B = μ₀nI, I = 2 A, and n = 100 turns/m, we can calculate the magnetic field strength B = (4π × 10⁻⁷) × 100 × 2 = 2.512 × 10⁻³ T. Then, we can calculate the torque τ = 0.5 × 2.512 × 10⁻³ × sin90° = 1.256 × 10⁻³ Nm.
⚠️ COMMON MISTAKE: Students often forget to take into account the angle between the magnetic moment and the magnetic field, which is crucial in calculating the torque experienced by the bar magnet.
11 Jun 26
🔗 More from Magnetism and Matter
Practice this chapter
Get AI-generated board exam questions, track your mastery, and identify weak spots.
Start Free →