Practice question
Question
A square loop of side \( 0.2 \, \text{m} \) with 30 turns carries \( 1.5 \, \text{A} \) in a magnetic
field of \( 0.4 \, \text{T} \). The plane of the loop is at \( 60^\circ \) to the field. What is the
torque?
Explanation
**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. Torque tau = N I A B sin θ , where A = 0.2 × 0.2 = 0.04 m² . tau = 30 × 1.5 × 0.04 × 0.4 × sin 60° = 1.8 × 0.4 × 0.866 = 0.6235 ≈ 0.62 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π
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