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
BETA
