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Electromagnetic Induction question

2003 · Shift 0 · Q125
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Electromagnetic Induction question

2003 · Shift 0 · Q125

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
When the current changes from +2A+ 2A+2A to −2A-2A−2A in 0.050.050.05 second, an e.m.f.e.m.f.e.m.f. of 8V8V8V is inducted in a coil. The coefficient of self- induction of the coil is
  1. A
    0.2H0.2H0.2H
  2. B
    0.4H0.4H0.4H
  3. C
    0.8H0.8H0.8H
  4. D
    0.1H0.1H0.1H
View written solutionFree

Correct answer: D

  1. Use the formula for self-induced emf

For a coil, the magnitude of induced emf is

e=L∣ΔIΔt∣e = L\left|\frac{\Delta I}{\Delta t}\right|e=L​ΔtΔI​​

where:

  • e=8 Ve = 8\,\text{V}e=8V
  • L=L =L= coefficient of self-induction
  • ΔI=\Delta I =ΔI= change in current
  • Δt=0.05 s\Delta t = 0.05\,\text{s}Δt=0.05s
  1. Find the change in current

The current changes from +2 A+2\,\text{A}+2A to −2 A-2\,\text{A}−2A.

So,

ΔI=−2−(+2)=−4 A\Delta I = -2 - (+2) = -4\,\text{A}ΔI=−2−(+2)=−4A

Hence,

∣ΔI∣=4 A|\Delta I| = 4\,\text{A}∣ΔI∣=4A

  1. Substitute into the formula

8=L(40.05)8 = L\left(\frac{4}{0.05}\right)8=L(0.054​)

8=L(80)8 = L(80)8=L(80)

  1. Solve for LLL

L=880=0.1 HL = \frac{8}{80} = 0.1\,\text{H}L=808​=0.1H

  1. Match with the options

L=0.1 HL = 0.1\,\text{H}L=0.1H

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

This matches the derived answer.

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