A circular coil of 60 turns and radius \( 8 \, \text{cm} \) carries a current of \( 0.9 \, \text{A} \). What is the magn
**Field at centre of circular loop** with N turns is B = μ₀ N I/(2R), R radius (m), direction along axis via right-hand rule, magnitude proportional to N I/R. For R = 0.09 m, N = 45, I = 1.2 A, B = 4π×10⁻⁷×45×1.2/(2×0.09) = 3.77×10⁻⁴ T, showing N enhancement. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 60 × 0.9/2 × 0.08) = (21.6 π × 10⁻⁶/0.16) = 1.35 π × 10⁻⁴ ≈ 4.24 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),
Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop
