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Electromagnetic Induction question

2019 · 10 Jan · Shift 2 · Q68
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Electromagnetic Induction question

2019 · 10 Jan · Shift 2 · Q68

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -
  1. A
    740 J
  2. B
    637.5 J
  3. C
    540 J
  4. D
    437.5 J
View written solutionFree

Correct answer: D

  1. Use self-induced emf formula

For an inductor, e=Ldidte = L\frac{di}{dt}e=Ldtdi​

Given:

  • Self-induced emf, e=25 Ve = 25\text{ V}e=25 V
  • Current changes from 10 A10\text{ A}10 A to 25 A25\text{ A}25 A in 1 s1\text{ s}1 s

So, didt=25−101=15 A/s\frac{di}{dt} = \frac{25-10}{1} = 15\text{ A/s}dtdi​=125−10​=15 A/s

Hence, L=edi/dt=2515=53 HL = \frac{e}{di/dt} = \frac{25}{15} = \frac{5}{3}\text{ H}L=di/dte​=1525​=35​ H

  1. Energy stored in an inductor

Energy stored is U=12LI2U = \frac{1}{2}LI^2U=21​LI2

Therefore, change in energy is ΔU=12L(I22−I12)\Delta U = \frac{1}{2}L\left(I_2^2 - I_1^2\right)ΔU=21​L(I22​−I12​)

Substitute: ΔU=12⋅53(252−102)\Delta U = \frac{1}{2}\cdot \frac{5}{3}\left(25^2 - 10^2\right)ΔU=21​⋅35​(252−102)

=56(625−100)= \frac{5}{6}(625-100)=65​(625−100)

=56⋅525= \frac{5}{6}\cdot 525=65​⋅525

=437.5 J= 437.5\text{ J}=437.5 J

  1. Match with options

The correct option is: 437.5 J\boxed{437.5\text{ J}}437.5 J​ which is Option D.

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