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AP EAMCET · PHYSICS · Electromagnetic Induction

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. A \(20 \times 10^{-6} \mathrm{~Wb}\)
  2. B \(5 \times 10^{-6} \mathrm{~Wb}\)
  3. C \(12 \times 10^{-6} \mathrm{~Wb}\)
  4. D \(10 \times 10^{-6} \mathrm{~Wb}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(10 \times 10^{-6} \mathrm{~Wb}\)

Step-by-step Solution

Detailed explanation

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