A square loop of side \( 0.1 \, \text{m} \) with 25 turns carries \( 2 \, \text{A} \) in a magnetic field of \( 0.8 \, \
**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. Torque tau = N I A B sin θ , where A = 0.1 × 0.1 = 0.01 m² . tau = 25 × 2 × 0.01 × 0.8 × sin 30° = 0.5 × 0.8 × 0.5 = 0.2 N m . Using F = q v B sinθ, F = I l B sinθ, B =
Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop