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QPages 230-234: Q36: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q36 Solution: A proton is projected with a velocity of perpendicular to a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly assume the acceleration is in the direction of the velocity or the magnetic field; however, the magnetic force $F = qvB$ is always perpendicular to both, meaning the acceleration is centripetal and changes only the direction of motion, not its speed.

A proton is projected with a velocity of 3×106 m s−13 \times 10^6 \text{ m s}^{-1} perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.

Step-by-Step Verified Solution

**Answer:** a=1.72×1014 m/s2a = 1.72 \times 10^{14} \text{ m/s}^2. **Step-BY-Step Explanation:** F=evB=ma⇒a=evBm=1.6×10−19×3×106×0.61.67×10−27=1.72×1014 m/s2F = e v B = m a \Rightarrow a = \frac{e v B}{m} = \frac{1.6 \times 10^{-19} \times 3 \times 10^6 \times 0.6}{1.67 \times 10^{-27}} = 1.72 \times 10^{14} \text{ m/s}^2.

All Questions in this Chapter

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