Verified Solution • 2025–26 Session moderate

QPages 306-313: Q64: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q64 Solution: A uniform magnetic field B exists in a cylindrical region,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently replace the circular loop length $2pi r$ with the diameter $2r$ when applying $oint\mathbf E\cdot d\mathbf l = -d\Phi/dt$, producing an $\pi$‑factor error in $E$. Remember that the induced electric field is tangential to the circular path, so the emf equals $E\times$(circumference $=2\pi r$), not $E\times$(diameter).

A uniform magnetic field B exists in a cylindrical region, shown dotted in figure. The magnetic field increases at a constant rate dB/dt. Consider a circle of radius r coaxial with the cylindrical region. a) Find the magnitude of the electric field E at a point on the circumference of the circle. b) Consider a point P on the side of the square circumscribing the circle. Show that the component of the induced electric field at P along ba is the same as the magnitude found in part a).

Step-by-Step Verified Solution

**Answer:** (a) E = (r/2)(dB/dt) (b) Component along ba equals E. **Step-by-Step Explanation:** (a) Emf induced in circle: _ = �r_(dB/dt). E _ 2�r = _, so E = (r/2)(dB/dt). (b) For the square, the flux is through area (2r)_. The component of E along ba equals the value from (a) [citation:5].

All Questions in this Chapter

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