Verified Solution • 2025–26 Session moderate

QPages 306-313: Q34: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q34 Solution: A circular copper-ring of radius r translates in its plane with …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the emf between two points as if it were the loop emf given by Faraday’s law, i.e. $$mathcal{E}= rac{dPhi}{dt},$$ and therefore look for the pair of points that encloses the largest area or the greatest flux change. In fact, for a moving conductor the motional emf between two points is $$mathcal{E}=(mathbf{v} imesmathbf{B})cdotmathbf{l},$$ where $mathbf{l}$ is the displacement vector from one point to the other. Thus the emf depends only on the component of $mathbf{l}$ perpendicular to $mathbf{v} imesmathbf{B}$, not on any enclosed area. The maximum occurs when $mathbf{l}$ is perpendicular to $mathbf{v}$ (i.e. the points are 90° apart in the direction of motion), and the minimum (zero) when $mathbf{l}$ is parallel to $mathbf{v}$.

A circular copper-ring of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametrically opposite points on the ring. a) Between which pair of points is the emf maximum? What is the value of this maximum emf? b) Between which pair of points is the emf minimum? What is the value of this minimum emf?

Step-by-Step Verified Solution

**Answer:** (a) Maximum: 2Brv (between points perpendicular to velocity). (b) Minimum: 0 (between points along velocity). **Step-by-Step Explanation:** The emf between diametrically opposite points is maximum when the line joining them is perpendicular to velocity, and zero when parallel [citation:3].

All Questions in this Chapter

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