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 $$4 \times {10^{ - 3}}Wb.$$ The self-inductance of the solenoid is
A.
$$3\,H$$
B.
$$2\,H$$
C.
$$1\,H$$
D.
$$4\,H$$
Answer :
$$1\,H$$
Solution :
Given, Number of turns of solenoid, $$N = 1000.$$
Current, $$I = 4A$$
Magnetic flux, $${\phi _B} = 4 \times {10^{ - 3}}Wb$$
$$\because $$ Self inductance of solenoid is given by
$$L = \frac{{{\phi _B}.N}}{I}\,.......\left( {\text{i}} \right)$$
Substitute the given values in Eq. (i), we get
$$L = \frac{{4 \times {{10}^{ - 3}} \times 1000}}{4} = 1\,H$$
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