Electromagnetic Induction - Result Question 24
24. 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
(a) 2.5 henry
(b) 2.0 henry
(c) 1.0 henry
(d) 40 henry
[2008]
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Answer:
Correct Answer: 24. (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 $N \phi=LI \Rightarrow L=\frac{Nf}{1}$
$=\frac{500^{\prime} 4^{\prime} 10^{-3}}{2}$ henry $=1 H$.
