The current passing through a coil of 120 turns and inductance. 40 mH is 30 mA . The magnetic flux linked…

The current passing through a coil of 120 turns and inductance. 40 mH is 30 mA . The magnetic flux linked with the coil is
  1. $20 \times 10^{-6} \mathrm{~Wb}$
  2. $5 \times 10^{-6} \mathrm{~Wb}$
  3. $12 \times 10^{-6} \mathrm{~Wb}$
  4. $10 \times 10^{-6} \mathrm{~Wb}$

Solution

$\mathrm{N}=120, \mathrm{~L}=40 \mathrm{mH}, \mathrm{I}=30 \mathrm{~mA}$
The magnetic flux linked with the coil is
$\phi=\frac{\mathrm{LI}}{\mathrm{~N}}=\frac{40 \times 10^{-3} \times 30 \times 10^{-3}}{120}=10 \times 10^{-6} \mathrm{~Wb}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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