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
- A \(20 \times 10^{-6} \mathrm{~Wb}\)
- B \(5 \times 10^{-6} \mathrm{~Wb}\)
- C \(12 \times 10^{-6} \mathrm{~Wb}\)
- D \(10 \times 10^{-6} \mathrm{~Wb}\)
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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