Given that the resultant mmf wave in the air gap of the three phase two pole induction motor is $$F(\theta , t) = \frac{3}{2}NI_m \cos(\omega t - \theta) \tag{1}$$ what's the value of the induced voltage in the phase coils? According to the Faraday's law of induction, we have $$\text{emf} = -N\frac{d\Phi}{dt}\tag{2}$$so we should calculate the flux linkage of the each phase coils and differentiate the result to obtain the induced voltage but I don't know how to find flux linkage. I've check various sources including Electric Machinery and Principles of Electric Machines and Power Electronics but they didn't help me to understand the solution. I'm mostly interested in the mathematical answer to this question rather than qualitative answer.

Edit: My problem main is as follows. According to the first equation, flux density should be of the form $$\vec{B} = B_{max}\cos(\omega t - \theta)\hat{r}$$so the flux linkage is $$\lambda = N\int_{-\pi/2}^{\pi/2}\int_{0}^{l}B_{max}\cos(\omega t - \theta)\hat{r}.rdzd\theta \hat{r} = NlrB_{max}\times2\cos(\omega t)$$the induced voltage should be $$\text{emf} = -\frac{d\lambda}{dt} = 2Nlr\omega B_{max}\sin(\omega t)$$ But it's given that $$e_a = k_w N_{ph} \frac{d\Phi_p}{dt} \cos(\omega_e t) - \omega_e k_w N_{ph} \Phi_p \sin(\omega_e t)$$ Apart from the scaling factors, why there should be $$\frac{d\Phi_p}{dt} \cos(\omega_e t)$$ term in the induced voltage? Also what changes has to made if the rotor also rotates? How the "slip" comes to the equations, in that case?

  • \$\begingroup\$ Perhaps amazon.com/… might help. \$\endgroup\$ – Charles Cowie Dec 28 '20 at 14:55
  • \$\begingroup\$ @CharlesCowie Thanks for your suggestion. Could you give an explanation about how those books derived induced voltage, please? I really tried to follow their reasoning but didn't understand how the formula for the induced voltage was derived. \$\endgroup\$ – S.H.W Dec 28 '20 at 15:15
  • \$\begingroup\$ I was hoping that the link I gave would do that. I believe that the flux linkage determination requires a detailed analysis of the machine construction details and finite element analysis. \$\endgroup\$ – Charles Cowie Dec 28 '20 at 16:31
  • \$\begingroup\$ @CharlesCowie You're right but I don't want to go that much. For example Fitzgerald gives $$e_a = k_w N_{ph} \frac{d\Phi_p}{dt} \cos(\omega_e t) - \omega_e k_w N_{ph} \Phi_p \sin(\omega_e t)$$ but I couldn't understand how this was obtained. \$\endgroup\$ – S.H.W Dec 28 '20 at 16:40
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    \$\begingroup\$ In the case of a 2-pole motor, the frequency fo the current is the same as the rotor speed. The second paragraph following the equation explains that one term is for the voltage due to the speed of the rotating field and the other is due to the frequency of the current. I don't think the slip has an effect on the voltage. It has an effect of the power conversion reflected in R2(1-s)/s. My understanding is pretty fuzzy. I don't think I can do better than that. \$\endgroup\$ – Charles Cowie Dec 28 '20 at 19:50

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