Solution:
Solution
Here, $$N = 400, L = 10 mH = 10 \, \times \, 10^{-3}H
I = 2 mA = 2 \, \times \, 10^{-3}A$$
total magnetic flux with the coil,
$$\phi = NLI = 400 \, \times \, (10 \, \times \, 10^{-3})\, \times \, 2 \, \times \, 10^{-3} = 8 \, \times \,10^{-3}SWb$$
Magnetic flux through the cross - section of the coil
= Magnetic flux linked with each turn
=$$\dfrac{\phi }{N} \, = \, \dfrac{8 \times 10^{-3}}{200} \, = \, 4 \times 10^{-5}Wb$$