Calculate initial back EMF (E1)

Using the formula for back EMF for a DC motor: $E_1 = V - I_{a1} R_a$.

$E_1 = 500 - (200 \times 0.12) = 500 - 24 = 476 \text{ V}$

Calculate final back EMF (E2)

Calculate the back EMF at the new armature current: $E_2 = V - I_{a2} R_a$.

$E_2 = 500 - (100 \times 0.12) = 500 - 12 = 488 \text{ V}$

Relate speed to back EMF and flux

Since $N \propto \frac{E}{\phi}$, we use the ratio: $\frac{N_2}{N_1} = \frac{E_2}{E_1} \times \frac{\phi_1}{\phi_2}$. Given $\phi_2 = 0.8 \phi_1$, the ratio $\frac{\phi_1}{\phi_2} = \frac{1}{0.8} = 1.25$.

$N_2 = N_1 \times \frac{E_2}{E_1} \times \frac{\phi_1}{\phi_2} = 250 \times \frac{488}{476} \times 1.25$

Final Calculation

Evaluating the expression to find the new speed $N_2$.

$N_2 = 250 \times 1.0252 \times 1.25 \approx 320.3 \text{ RPM}$