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QPages 230-234: Q37: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q37 Solution: a) An electron moves along a circle of radius 1 m in a…

⚠️ Common Exam Traps & Conceptual Pitfalls

The formula $v=qBr/m$ is valid only for non‑relativistic speeds. If the calculation yields $v>c$, the assumption breaks down and the particle’s speed must be found using relativistic dynamics or the result is simply not physically possible.

a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T. What would be its speed? Is it reasonable? b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed?

Step-by-Step Verified Solution

**Answer:** (a) ve=8.8×1010 m/sv_e = 8.8 \times 10^{10} \text{ m/s}, not reasonable. (b) vp=4.8×107 m/sv_p = 4.8 \times 10^7 \text{ m/s}. **Step-by-Step Explanation:** (a) ve=eBrme=1.6×10−19×0.5×19.1×10−31=8.8×1010 m/sv_e = \frac{e B r}{m_e} = \frac{1.6 \times 10^{-19} \times 0.5 \times 1}{9.1 \times 10^{-31}} = 8.8 \times 10^{10} \text{ m/s}. This exceeds cc, so it is not reasonable. (b) vp=eBrmp=1.6×10−19×0.5×11.67×10−27=4.8×107 m/sv_p = \frac{e B r}{m_p} = \frac{1.6 \times 10^{-19} \times 0.5 \times 1}{1.67 \times 10^{-27}} = 4.8 \times 10^7 \text{ m/s}.

All Questions in this Chapter

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