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Electromagnetic Induction question

2019 · 9 Apr · Shift 2 · Q70
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Electromagnetic Induction question

2019 · 9 Apr · Shift 2 · Q70

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic flux of 10–3 Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' and a current of 2 A passes through 'Q', the flux through 'P' is :-
  1. A
    3.67 × 10–4 Wb
  2. B
    3.67 × 10–3 Wb
  3. C
    6.67 × 10–4 Wb
  4. D
    6.67 × 10–3 Wb
View written solutionFree

Correct answer: C

  1. Use the reciprocity of mutual inductance

For two coils PPP and QQQ, the mutual inductance is the same in either direction:

MPQ=MQPM_{PQ} = M_{QP}MPQ​=MQP​

Also,

M=flux linkagecurrent producing itM = \frac{\text{flux linkage}}{\text{current producing it}}M=current producing itflux linkage​

Since number of turns is not mentioned, we interpret the given flux directly for mutual relation.

  1. Case 1: Current in coil PPP produces flux through QQQ

Given:

  • Current in PPP: IP=3 AI_P = 3\,\text{A}IP​=3A
  • Flux through QQQ: ϕQ=10−3 Wb\phi_Q = 10^{-3}\,\text{Wb}ϕQ​=10−3Wb

So,

M=ϕQIP=10−33 HM = \frac{\phi_Q}{I_P} = \frac{10^{-3}}{3}\,\text{H}M=IP​ϕQ​​=310−3​H

  1. Case 2: Current in coil QQQ produces flux through PPP

Given:

  • Current in QQQ: IQ=2 AI_Q = 2\,\text{A}IQ​=2A

Required flux through PPP, say ϕP\phi_PϕP​.

Using the same mutual inductance,

M=ϕPIQM = \frac{\phi_P}{I_Q}M=IQ​ϕP​​

Thus,

ϕP=MIQ=(10−33)(2)\phi_P = M I_Q = \left(\frac{10^{-3}}{3}\right)(2)ϕP​=MIQ​=(310−3​)(2)

ϕP=23×10−3\phi_P = \frac{2}{3}\times 10^{-3}ϕP​=32​×10−3

ϕP=6.67×10−4 Wb\phi_P = 6.67 \times 10^{-4}\,\text{Wb}ϕP​=6.67×10−4Wb

  1. Match with options

6.67×10−4 Wb6.67 \times 10^{-4}\,\text{Wb}6.67×10−4Wb

So the correct option is C.

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