alleryA current of 4 mA in a coil of self-inductance 10 mH and 200 turns produces total flux:
A$8 \times 10^{-3}$ Wb total
B$4 \times 10^{-5}$ Wb total (i.e. $2 \times 10^{-7}$ Wb per turn)
C$2 \times 10^{-5}$ Wb total
D$2 \times 10^{-7}$ Wb per turn
Answer & Solution
Correct answer: B. $4 \times 10^{-5}$ Wb total (i.e. $2 \times 10^{-7}$ Wb per turn)
**Total flux linkage:** $\phi = L i = (10 \times 10^{-3}) \times (4 \times 10^{-3}) = 4 \times 10^{-5}$ Wb.
Flux per turn = $\phi/N = (4 \times 10^{-5})/200 = 2 \times 10^{-7}$ Wb.
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