1. **Problem statement:** An iron ring has a cross-sectional area of 3 cm² and a mean diameter of 25 cm. An air gap of 0.4 mm is cut across the ring. The ring is wound with a coil of 200 turns carrying a current of 2 A. The total magnetic flux is 0.24 mWb. Find the relative permeability of iron assuming no magnetic leakage. 2. **Given data:** - Cross-sectional area, $A = 3\ \text{cm}^2 = 3 \times 10^{-4}\ \text{m}^2$ - Mean diameter, $d = 25\ \text{cm} = 0.25\ \text{m}$ - Air gap length, $l_g = 0.4\ \text{mm} = 0.0004\ \text{m}$ - Number of turns, $N = 200$ - Current, $I = 2\ \text{A}$ - Magnetic flux, $\Phi = 0.24\ \text{mWb} = 0.24 \times 10^{-3} = 2.4 \times 10^{-4}\ \text{Wb}$ 3. **Formulas and concepts:** - Magnetic field strength, $H = \frac{NI}{l}$ where $l$ is the length of the magnetic path. - Magnetic flux density, $B = \frac{\Phi}{A}$ - Magnetic field intensity in iron and air gap add up: $NI = H_{iron} l_{iron} + H_{gap} l_g$ - $B$ is continuous across iron and air gap, so $B = \mu_0 \mu_r H_{iron} = \mu_0 H_{gap}$ - $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$ (permeability of free space) 4. **Calculate magnetic path length of iron:** $$l_{iron} = \pi d = \pi \times 0.25 = 0.7854\ \text{m}$$ 5. **Calculate magnetic flux density $B$:** $$B = \frac{\Phi}{A} = \frac{2.4 \times 10^{-4}}{3 \times 10^{-4}} = 0.8\ \text{T}$$ 6. **Calculate magnetic field intensity in air gap $H_{gap}$:** $$H_{gap} = \frac{B}{\mu_0} = \frac{0.8}{4\pi \times 10^{-7}} = \frac{0.8}{1.2566 \times 10^{-6}} \approx 636619.8\ \text{A/m}$$ 7. **Calculate magnetic field intensity in iron $H_{iron}$:** $$H_{iron} = \frac{B}{\mu_0 \mu_r} = \frac{0.8}{4\pi \times 10^{-7} \mu_r} = \frac{636619.8}{\mu_r}$$ 8. **Apply Ampere's law:** $$NI = H_{iron} l_{iron} + H_{gap} l_g$$ Substitute values: $$200 \times 2 = \frac{636619.8}{\mu_r} \times 0.7854 + 636619.8 \times 0.0004$$ $$400 = \frac{636619.8 \times 0.7854}{\mu_r} + 254.6$$ 9. **Solve for $\mu_r$:** $$400 - 254.6 = \frac{500000}{\mu_r}$$ $$145.4 = \frac{500000}{\mu_r}$$ $$\mu_r = \frac{500000}{145.4} \approx 3437$$ **Final answer:** The relative permeability of iron is approximately **3437**.