Verified Solution • 2025–26 Session moderate

QPages 304-305: Q1: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 304-305: Q1 Solution: A rod of length l rotates with a small but uniform angular…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the full rod length $l$ into the standard motional emf formula $\frac{1}{2}B\omega L^2$, yielding the incorrect option $\frac{1}{2}\omega B l^2$. Since the rod rotates about its center, the effective length $L$ for the segment from the center to the end is only $l/2$, making the correct potential difference $\frac{1}{2}B\omega(l/2)^2 = \frac{1}{8}\omega B l^2$.

A rod of length l rotates with a small but uniform angular velocity ω\omega about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is _________ .

Multiple Choice Options

  • A Zero
  • B 18ωBl2\frac{1}{8}\omega B l^2
  • C 12ωBl2\frac{1}{2}\omega B l^2
  • D Bωl2B\omega l^2

Step-by-Step Verified Solution

**Answer:** (b) (1/8)_Bl_. **Step-by-Step Explanation:** The emf between the centre and end of a rotating rod is (1/2)B_(l/2)_ = (1/8)B_l_.

All Questions in this Chapter

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