Electromagnetic Induction Ques 17
17. 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 $4 \times 10^{-3}$ $ Wb$. The self-inductance of the solenoid is
[2008]
(a) $2.5$ henry
(b) $ 2.0\ \text{H}$
(c) $1.0$ henry
(d) $40$ henry
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Answer:
Correct Answer: 17.(c)
Solution:
- (c) Total number of turns in the solenoid, $N=500$
Current, $I=2$ $ A$.
Magnetic flux linked with each turn
$=4 \times 10^{-3}$ $ Wb$
As, $\phi=LI$ or $N \phi=LI \Rightarrow L=\frac{N\phi}{I}$
$=\frac{500\cdot 4\cdot 10^{-3}}{2}$
henry $=1\ \text{H}$.
BETA
