Verified Solution • 2025–26 Session moderate

QPages 306-313: Q76: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q76 Solution: What are the values of the self-induced emf in the circuit of…

⚠️ Common Exam Traps & Conceptual Pitfalls

When computing the self‑induced emf, students often drop the negative sign in $E_{ ext{self}}=-L, rac{di}{dt}$. The minus sign reflects Lenz’s law: the induced emf opposes the change in current. Forgetting it gives the wrong polarity and can lead to incorrect conclusions about the direction of the induced voltage. Always keep the $-$ sign and check that the sign of $di/dt$ matches the direction of the current change.

What are the values of the self-induced emf in the circuit of the previous problem at the times indicated therein?

Step-by-Step Verified Solution

**Answer:** E_self = -L(di/dt). **Step-by-Step Explanation:** The self-induced emf is E = -L(di/dt). At the times given in the previous problem, this is calculated from the di/dt values [citation:4].

All Questions in this Chapter

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