--- title: "Iron ring with saw cut, 4 mWb flux" subject: "Electrical Machines" chapter: 1 key: "semester_4/electrical_machines:ring-saw-cut-30cm-4mwb" source: https://www.bctengineer.com/q/semester_4/electrical_machines/ring-saw-cut-30cm-4mwb site: BCT Engineering --- # Iron ring with saw cut, 4 mWb flux Electrical Machines · Chapter 1: Magnetic Circuits and Induction - Asked in: 2072 Asoj; 2070 Magh, 8 marks (asked 2 times) - Video: [Iron ring | Electromagnetism | Magnetic Circuit | Problem 1](https://www.youtube.com/watch?v=bjzbJLAnf4M) — Aryabhat Institute - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/ring-saw-cut-30cm-4mwb - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1 ## Question A ring of 30 cm mean diameter is made up of round iron rod 2.5 cm in diameter. A saw cut of 1 mm is made on the ring. It is uniformly wound with 500 turns of wire. Calculate the current required by the exciting coil to produce a total flux of 4 mWb. Assume a relative permeability of iron at this flux density as 800. Neglect leakage and fringing. ## Short answer Compute iron path length (0.9415 m) and cross-sectional area (4.909×10^-4 m^2). Total reluctance = iron + gap = 3.529×10^6 AT/Wb. Current = flux × reluctance / N = 4×10^-3 × 3.529×10^6 / 500 = 28.2 A. ## Answer The exciting current is found from the total mmf, which is the sum of the mmf for the iron path and for the air gap, using the reluctance of each part. **Given:** mean diameter $D = 30$ cm, rod diameter $d = 2.5$ cm, gap $l_g = 1$ mm, $N = 500$, $\Phi = 4$ mWb, $\mu_r = 800$. **Dimensions** $$ \begin{aligned} l_{total} &= \pi D = \pi \times 0.30 = 0.9425\ \text{m} \\ l_i &= 0.9425 - 0.001 = 0.9415\ \text{m} \\ A &= \frac{\pi}{4}(0.025)^2 = 4.909\times10^{-4}\ \text{m}^2 \end{aligned} $$ **Reluctances** $$ \begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{0.9415}{4\pi\times10^{-7}\times800\times4.909\times10^{-4}} = 1.908\times10^{6}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.001}{4\pi\times10^{-7}\times4.909\times10^{-4}} = 1.621\times10^{6}\ \text{AT/Wb} \end{aligned} $$ The two reluctances are in series (the same flux passes through both), so $S = S_i + S_g = 3.529\times10^{6}$ AT/Wb. **Mmf and current** $$ \begin{aligned} NI &= \Phi S = 4\times10^{-3}\times3.529\times10^{6} = 14\,116\ \text{AT} \\ I &= \frac{14\,116}{500} = 28.23\ \text{A} \end{aligned} $$ The flux density is $B = \Phi/A = 8.15$ T, which is far above the saturation of real iron, so the given $\mu_r = 800$ is only an assumption of the problem. **Answer: $I \approx 28.2$ A** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the iron reluctance and the air-gap reluctance added to get the total reluctance? A. The same flux passes through both parts, so they are in series B. They carry different fluxes, so they are in parallel C. Air has no reluctance, so only iron counts D. The gap is added only when leakage is present Answer: A. The flux passes through the iron and then the gap, so the reluctances are in series and S = S_i + S_g. (Key point: Series reluctances) 2. Which expression gives the reluctance of the 1 mm saw cut? A. l_g / (mu_0 mu_r A) B. l_g / (mu_0 A) C. mu_0 A / l_g D. l_g x A / mu_0 Answer: B. Air has mu_r = 1, so S_g = l_g/(mu_0 A). (Key point: Gap reluctance formula) 3. After finding S = 3.529 x 10^6 AT/Wb, how is the exciting current obtained? A. I = S / (N Phi) B. I = Phi S / N C. I = N / (Phi S) D. I = Phi / (N S) Answer: B. NI = Phi S, so I = Phi S / N = 14116/500 = 28.2 A. (Key point: Current from total mmf) 4. In this ring the 1 mm gap has a reluctance comparable to the 0.94 m of iron. Why? A. The gap has a larger area than the iron B. Air has mu_r = 1 while iron is taken as 800 C. The flux is larger in the gap D. The gap carries all the leakage flux Answer: B. Reluctance is inversely proportional to permeability, so a short air path equals a long iron path. 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