Problem 1: Magnetic Field at Centre and Axis of a Circular Coil

Given: N = 100 turns, R = 10 cm = 0.10 m, I = 2 A, μ_{0} = 4π × 10^{-7} T·m/A

At centre (x = 0): $B_{\text{centre}} = \frac{\mu_0 NI}{2R} = \frac{(4\pi \times 10^{-7}\ \text{T·m/A})(100)(2\ \text{A})}{2 \times 0.10\ \text{m}}$

Numerator: $4\pi \times 10^{-7}\ \text{T·m/A} \times 200\ \text{A} = 800\pi \times 10^{-7}\ \text{T·m}$

Denominator: $0.20\ \text{m}$

$B_{\text{centre}} = \frac{800\pi \times 10^{-7}\ \text{T·m}}{0.20\ \text{m}} = 4\pi \times 10^{-4}\ \text{T} = 1.257\ \text{mT}$

At axial point x = 10 cm = R = 0.10 m: $B_{\text{axis}} = \frac{\mu_0 NI R^2}{2(R^2 + x^2)^{3/2}} = \frac{(4\pi \times 10^{-7})(100)(2)(0.10)^2}{2(0.01 + 0.01)^{3/2}\ \text{m}^3}$

$(0.02)^{3/2} = 0.02 \times \sqrt{0.02} = 0.02 \times 0.1414\ \text{m}^{3/2} = 2.828 \times 10^{-3}\ \text{m}^{3/2}$

Numerator: $4\pi \times 10^{-7} \times 200 \times 0.01 = 8\pi \times 10^{-7}\ \text{T·m}^3/\text{A} \times \text{A} = 8\pi \times 10^{-7}\ \text{T·m}^3$

Denominator: $2 \times 2.828 \times 10^{-3}\ \text{m}^3 = 5.657 \times 10^{-3}\ \text{m}^3$

$B_{\text{axis}} = \frac{8\pi \times 10^{-7}}{5.657 \times 10^{-3}}\ \text{T} = 4.44 \times 10^{-4}\ \text{T} = 0.444\ \text{mT}$

Verification: $B_{\text{axis}} = B_{\text{centre}} / 2\sqrt{2} = 1.257/2.828 = 0.444\ \text{mT}$ ✓