A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 1 0^{- 3} ω b$ . The selfinductance of the solenoid isNEET 2008 (Mains)
Options:
Answer:
D
Solution:
$Nϕ = L i 500 \times 4 \times 1 0^{- 3} = 2 L L = 1.0 henry$

Question

A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 1 0^{- 3} ω b$ . The selfinductance of the solenoid isNEET 2008 (Mains)

Options:

Answer:

D

Solution:

$Nϕ = L i 500 \times 4 \times 1 0^{- 3} = 2 L L = 1.0 henry$

TestHub · Accepted Answer

​ Nϕ = L i 500 × 4 × 1 0 − 3 = 2 L L = 1.0 henry ​

Physics - EMI/AC Question with Solution | TestHub

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