SATHEE: Electro Magnetic Induction And Alternating Currents Question 498

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 $4 \times 10^{- 3}, W b$ . 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 $= 4 \times 10^{- 3}, W b$

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

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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 4×10−3,Wb . The self- inductance of the solenoid is

Options:

Answer:

Solution:

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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 $4 \times 10^{- 3}, W b$ . The self- inductance of the solenoid is