Answer
a. $M=1.96H $.
b. $\Phi_{B1}=7.11\times10^{-3}Wb$.
Work Step by Step
a. $M=\frac{ N_2\Phi_{B2} }{i_1}=\frac{400(0.0320Wb)}{6.52A}=1.96H $.
b. Also, $M=\frac{ N_1\Phi_{B1} }{i_2}$. Solve for the average flux per turn in coil 1.
$$\Phi_{B1}=\frac{Mi_2}{N_1}=\frac{(1.96H)(2.54A)}{700}=7.11\times10^{-3}Wb$$
