Verified Solution • 2025–26 Session moderate

QPages 306-313: Q59: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q59 Solution: Figure shows a conducting circular loop of radius a placed in a …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently use the linear motional emf formula $\varepsilon = B\,a\,\omega$ for the rotating rod, ignoring that points along the rod have speeds $v=\omega r$ that vary with radius; the correct emf is $\varepsilon = \frac12 B a^{2}\omega$. Remember to integrate $vB\,dr$ over the rod length, which introduces the essential ½ factor (and the $\omega$) before computing current and force.

Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity ω\omega. Find the force.

Step-by-Step Verified Solution

**Answer:** F = (B__a_)/(2R). **Step-Step Explanation:** The rod rotates, inducing emf _ = _B_a_. Current i = _/R. Force needed at midpoint: F _ (a/2) = i _ a _ B _ (a/2) [citation:3].

All Questions in this Chapter

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