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Question

Three solenoid coils of the same dimension, the same number of turns and the same number of layers of winding are taken. Coil 1 with inductanceL1 was wound using a Mn wire of resistance 11m-1, coil 2with inductanceL2 was wound using a similar wire but the direction of winding was reversed in each layer, coil 3 with inductance L3 was wound using a superconducting wire. The self-inductance of the coils L1,L2andL3 are


  1. L1=L2=L3

  2. L1=L2=L3=0

  3. L1=L2,L3=0

  4. L1>L2>L3

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Solution

The correct option is C

L1=L2,L3=0


Step 1: Given information

Mn wire of resistance 11m-1

The self-inductance coil 1, coil 2 and coil 3= L1,L2andL3

Step 2: Formula

Φ=B.AwhereΦismagneticfluxdensity,Bismagneticfeild

Φ=μ0NAi/l

Step 3: Calculation

The self-inductance of a coil,

L=NΦiL=μ0NAili

L=μ0NAl

L=μ0NAllxl

L=μ0Nl2Al

L=μ0n2AlifNl=n

Where,

  1. μ0 = permeability of air
  2. n = number of turns per unit length
  3. l= length of the solenoid
  4. A = area of cross-section
  5. i = current

Explanation

  1. The coil's self-inductance is affected by its cross-sectional area, length, and number of turns. It will not be affected by the material utilized.
  2. The current flow will be continuous if the superconductor is below the critical temperature.
  3. The inductance will no longer have the property of inductance. As a result,L1=L2,L3=0.

Hence, the correct option is (B).


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