A circular current loop of radius 2 cm carries current 100√2 A. Its unit normal is n̂ = (î+k̂) ÷ √2. The loop is placed in the magnetic field B = B₀(3î+2k̂), where B₀ = 4×10⁻³ T. Taking π = 3.14, what torque vector acts on the loop?
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Question

A circular current loop of radius 2 cm carries current 100√2 A. Its unit normal is n̂ = (î+k̂) ÷ √2. The loop is placed in the magnetic field B = B₀(3î+2k̂), where B₀ = 4×10⁻³ T. Taking π = 3.14, what torque vector acts on the loop?

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G1 Physics · Accepted Answer

Magnetic dipole moment of the loop is M = IA n̂. The loop area is A = πr² = 3.14×(2×10⁻²)² = 3.14×4×10⁻⁴ m². Therefore, M = 100√2 × 3.14×4×10⁻⁴ × [(î+k̂) ÷ √2]. The √2 factors cancel: M = 100×3.14×4×10⁻⁴(î+k̂) = 0.1256(î+k̂) A·m². The magnetic field is B = 4×10⁻³(3î+2k̂). Torque is τ = M×B. Thus, τ = 0.1256×4×10⁻³ [(î+k̂)×(3î+2k̂)]. Now, î×3î = 0, î×2k̂ = -2ĵ, k̂×3î = 3ĵ, k̂×2k̂ = 0. Therefore, (î+k̂)×(3î+2k̂) = ĵ. Hence, τ = 0.1256×4×10⁻³ ĵ = 5.024×10⁻⁴ ĵ N·m. Write this as 5.024×10⁻⁴ = 5024×1…