Verified Solution • 2025–26 Session moderate

QPages 249-253: Q53: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q53 Solution: Consider the situation of the previous problem. A particle…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often set $qvB = mv^2/r$ and then solve for $v$ as $v = \sqrt{\frac{qBr}{m}}$, mistakenly squaring both sides; the correct step is to divide both sides by $v$ (since $v\neq0$) yielding $v = \frac{qBr}{m}$.

Consider the situation of the previous problem. A particle having charge qq and mass mm is projected from the point QQ in a direction going into the plane of the diagram. It is found to describe a circle of radius rr between the two plates. Find the speed of the charged particle.

Step-by-Step Verified Solution

**Answer:** v=qBrmv = \frac{qBr}{m}. **Step-by-Step Explanation:** The particle moves in a circle of radius rr between the plates. The magnetic field is due to the surface currents. v=qBrmv = \frac{qBr}{m} where BB is the field between the plates.

All Questions in this Chapter

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