Q.

A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4×10-3 Wb. The self-inductance of the solenoid is

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a

2H

b

1H

c

4H

d

3H

answer is B.

(Detailed Solution Below)

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Detailed Solution

Here, N=1000,I=4 A,ϕ0=4×10-3 Wb Total flux linked with the solenoid, ϕ=Nϕ0=1000×4×10-3 Wb=4 Wb Since, ϕ=LI∴ Self-inductance of solenoid, L=ϕI=4 Wb4 A=1H
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