A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4×10-3 Wb. The self-inductance of the solenoid is

A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{- 3} Wb$ . The self-inductance of the solenoid is

A
2H
B
1H
C
4H
D
3H

Solution:

$Here, N = 1000, I = 4 A, ϕ_{0} = 4 \times 10^{- 3} Wb$

$Total flux linked with the solenoid, ϕ = N ϕ_{0}$

$= 1000 \times 4 \times 10^{- 3} Wb = 4 Wb$

$Since, ϕ = L I$

$∴ Self-inductance of solenoid, L = \frac{ϕ}{I} = \frac{4 Wb}{4 A} = 1 H$

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