Question 17: Units , Dimensions and Error Analysis

Question

A long solenoid has 400 turns. When a current of 100 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 mWb . The self-inductance of the solenoid is

Accepted Answer

First, recall that the self-inductance $L$ is given by
$$ L \;=\;\frac{N\,\Phi}{I} $$
where
$N$ = number of turns
$\Phi$ = flux through each turn
$I$ = current
Here:
$N = 400$
$\Phi = 4	ext{ mWb} = 4	imes10^{-3}	ext{ Wb}$
$I = 100	ext{ A}$
Compute the total flux linkage $N\Phi$:
$$ N\Phi = 400 	imes 4	imes10^{-3} = 1.6	ext{ WbÃÂ·turn} $$
Then
$$ L = \frac{1.6}{100} = 0.016	ext{ H} = 16	ext{ mH} $$
So the correct answer is
16 mH
(Option B).

Answer

1.6 mH

Answer

16 mH

Answer

16 H

Answer

0.16 mH

Units , Dimensions and Error Analysis

A long solenoid has 400 turns. When a current of 100 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 mWb . The self-inductance of the solenoid is

Choose the correct answer: