2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electro Magnetic Induction

JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Self Evaluation Test - Electromagnetic Induction

  • question_answer
    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

    A) 2.5 henry

    B) 2.0 henry

    C) 1.0 henry

    D) 40 henry

    Correct Answer: C

    Solution :

    [c] 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


You need to login to perform this action.
You will be redirected in 3 sec spinner