Two identical circular wires of radius 20 cm and carrying current √2 A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires i $s ______$ × 10⁻⁸ T.

Question

G1 Physics · Accepted Answer

Magnetic field at the centre of one N-turn circular loop is B = μ₀NI/(2R) Here each loop has: N = 1 I = √2 A R = 0.20 m So field due to each loop: B₀ = (4π×10⁻⁷)(√2)/(2×0.20) The two loops are in mutually perpendicular planes, so their fields at the common centre are perpendicular. Net field: B(net) = √(B₀² + B₀²) = √2 B₀ Thus, B(net) = μ₀I√2/(2R) = (4π×10⁻⁷)(2)/(0.40) = 2π×10⁻⁶ T ≈ 6.28×10⁻⁶ T = 628×10⁻⁸ T Final value: 628

Two identical circular wires of radius 20 cm and carrying current √2 A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires i s ______ × 10⁻⁸ T.