DC motor induced back EMF
The conductor (not the whole coil, the part of coil facing the pole) is moving in a magnetic field which is Faraday's motional emf expression:-
$$E = -B.l.v.sin\theta$$
B = flux linked with the conductor
l = distance of the conductor from the pole (air-gap length)
v = velocity of the conductor
theta = angle b/w velocity and  magnetic field vectors
(always 90 degree here)

$$E = -B.l.v$$
$$E = -B.l.\frac{d}{2}.\omega$$
Total back emf is given adding both EMFs in conductor facing North Pole and conductor facing South Pole
$$E_{b} = -B.l.d.\omega$$
$$E_{b} = -B.l.d.2\pi n$$

Also in the flux waveform the constant flux time should be more (B1-B2, R1-R2) and similarly for back emf waveform 0 back emf part should be more (B1-B2, R1-R2).