7. A generator has a no load emf of 125 V and has 175 armature conductors. If a gasoline engine rotates the armature of the generator at a speed of \( 3500 \mathrm{r} / \mathrm{m} \), find the strength of the field flux per pole in Webers.

To find the strength of the field flux per pole in Webers, we can use the formula for the induced emf in a generator, which is: \[ E = \frac{N \cdot \Phi \cdot P \cdot n}{60} \] Where: - \(E\) is the no-load emf (125 V), - \(N\) is the number of armature conductors (175), - \(\Phi\) is the flux per pole in Webers, - \(P\) is the number of poles (assuming a typical generator with 2 poles unless stated otherwise), - \(n\) is the speed of rotation in revolutions per minute (3500 r/m). Assuming \(P = 2\), we can rearrange the equation to solve for \(\Phi\): \[ \Phi = \frac{E \cdot 60}{N \cdot P \cdot n} \] Now substituting the values into the equation: \[ \Phi = \frac{125 \cdot 60}{175 \cdot 2 \cdot 3500} \] Calculating this gives: \[ \Phi \approx \frac{7500}{1225000} \approx 0.00612 \text{ Wb} \] So, the strength of the field flux per pole is approximately **0.00612 Webers**.