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Electromagnetic Induction question

2019 · 9 Apr · Shift 1 · Q59
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Electromagnetic Induction question

2019 · 9 Apr · Shift 1 · Q59

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to :
  1. A
    1/L2
  2. B
    1/L
  3. C
    L
  4. D
    L2
View written solutionFree

Correct answer: B

  1. For a long solenoid, the inductance is

Ls=μ0μrN2AℓL_s = \mu_0 \mu_r \frac{N^2 A}{\ell}Ls​=μ0​μr​ℓN2A​

where:

  • LsL_sLs​ = inductance of the solenoid
  • NNN = total number of turns
  • AAA = cross-sectional area
  • ℓ\ellℓ = length of the solenoid
  1. According to the question:
  • total number of turns NNN is fixed
  • cross-sectional area AAA is fixed
  • only the length ℓ\ellℓ is varied

So from the formula,

Ls∝1ℓL_s \propto \frac{1}{\ell}Ls​∝ℓ1​

  1. Replacing ℓ\ellℓ by the given symbol LLL for length,

Inductance∝1L\text{Inductance} \propto \frac{1}{L}Inductance∝L1​

  1. Now check the options:
  • A: 1L2\frac{1}{L^2}L21​ — incorrect
  • B: 1L\frac{1}{L}L1​ — correct
  • C: LLL — incorrect
  • D: L2L^2L2 — incorrect

Therefore, the correct option is B.

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