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Question: Answered & Verified by Expert
The total emf induced in a closely would coil of $\mathrm{N}$ turns in which the magnetic flux linked with the coil changing at the rate $\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$ is
PhysicsElectromagnetic InductionAP EAMCETAP EAMCET 2023 (17 May Shift 2)
Options:
  • A $-\mathrm{N} \frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$
  • B $\mathrm{N} \frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$
  • C $-\mathrm{N} \frac{\mathrm{d}^2 \phi_{\mathrm{B}}}{\mathrm{dt}^2}$
  • D $-\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$
Solution:
1753 Upvotes Verified Answer
The correct answer is: $-\mathrm{N} \frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$
For a coil of $\mathrm{N}$ turns, the emf across the whole coil is $\varepsilon=-\mathrm{N} \frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$

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