A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 x 10-3 Wb. The self-inductance of the solenoid is: Option 1) 4 H Option 2) 3 H Option 3) 2 H Option 4) 1 H

As discussed in

Co efficient of self induction -

$\phi \, \alpha \, I\Rightarrow N\phi \, \alpha\: I$

$N\phi \,=L\, I$

$L=\frac{N\phi }{I}\,$

- wherein

$N\phi =$ Number of flux linkage with coil.

$N=1000, \:\:\:\:\:I=4A, \:\:\:\:\:\phi_{o}=4 \times 10^{-3}wb$

Total flux linked with the solenoid, $\phi =N \phi _o$$\phi=1000 \times 4 \times 10^{-3}wb=4wb$

$\phi=LI\:\:\:\:\:\Rightarrow L=\frac{\phi}{I}=\frac{4wb}{4A}=1H$

Option 1)

4 H

This option is incorrect.

Option 2)

3 H

This option is incorrect.

Option 3)

2 H

This option is incorrect.

Option 4)

1 H

This option is correct.

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