Electro Magnetic Induction And Alternating Currents Question 498

Question: 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

Options:

A) 2.5 henry

B) 2.0 henry

C) 1.0 henry

D) 40 henry

Show Answer

Answer:

Correct Answer: C

Solution:

  • Total number of turns in the solenoid, N=500 Current, I =2A.

    Magnetic flux linked with each turn $ =4\times {{10}^{-3}},Wb $

    As, $ \phi =LI $ or $ N\phi =LI $
    $ \Rightarrow ,L=\frac{N\phi }{1}=\frac{500\times 4\times {{10}^{-3}}}{2} $ henry =1 H



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