A 200 turn coil of self inductance 20 mH carries a current of 4 mA. The magnetic flux linked with each turn of the coil would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $4 × 10^{-7}\, Wb$
Given:
Number of turns, $N = 200$
Self-inductance, $L = 20 \ \text{mH} = 20 \times 10^{-3} \ \text{H}$
Current, $I = 4 \ \text{mA} = 4 \times 10^{-3} \ \text{A}$
Flux linked with the coil: $\Phi_\text{total} = L I$
Magnetic flux linked with each turn: $\phi = \frac{\Phi_\text{total}}{N} = \frac{L I}{N}$
$\phi = \frac{(20 \times 10^{-3}) \cdot (4 \times 10^{-3})}{200} = \frac{8 \times 10^{-5}}{200} = 4 \times 10^{-7} \ \text{Wb}$
Magnetic flux linked with each turn: $\phi = 4 \times 10^{-7} \ \text{Wb}$