-->

Electric Charges And Fields

Question
CBSEENPH12039785

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 4x10-3 Wb. The self inductance of the solenoid is,

  • 3H

  • 2H

  • 1H

  • 4H

Solution

C.

1H

Given,
Number of turns in the solenoid, N= 1000
Current, I = 4 A
Magnetic flux, straight phi subscript straight B = 4 x 10-3 Wb
Self-inductance of solenoid is given by,
straight L space equals space fraction numerator straight phi subscript straight B. straight N over denominator straight I end fraction     ... (i)
Putting the value of equation in (i), we get
L = fraction numerator 4 straight x 10 to the power of negative 3 end exponent straight x 1000 over denominator 4 end fraction
  = 1 H