Exercises
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
Given: N = 100 turns, R = 8.0 cm = 0.08 m, I = 0.40 A
Formula for magnetic field at the centre of a circular coil:
B = µ₀NI / (2R)
B = (4π × 10⁻⁷ × 100 × 0.40) / (2 × 0.08)
B = (4π × 10⁻⁷ × 40) / 0.16
B = (160π × 10⁻⁷) / 0.16
B = 1000π × 10⁻⁷ T
B = π × 10⁻⁴ T ≈ 3.14 × 10⁻⁴ T
Answer: B ≈ 3.14 × 10⁻⁴ T (= π × 10⁻⁴ T)
