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
Answer :
1.0 henry
Solution :
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, $$\phi = LI\,{\text{or}}\,N\phi = LI$$
$$ \Rightarrow L = \frac{{N\phi }}{1} = \frac{{500 \times 4 \times {{10}^{ - 3}}}}{2}{\text{henry}} = 1\,H.$$
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