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QPages 230-234: Q47: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q47 Solution: A particle moves in a circle of diameter 1.0 cm under the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the given diameter directly into the radius formula $r = mv/qB$ without halving it, leading to a result that is exactly double the correct value. Always verify that the geometric parameter provided matches the specific variable required by the kinematic or dynamic equation.

A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V m−1200 \text{ V m}^{-1} makes the path straight. Find the charge/mass ratio of the particle.

Step-by-Step Verified Solution

**Answer:** q/m=50 C/kgq/m = 50 \text{ C/kg}. **Step-by-Step Explanation:** For the path to be straight, qE=qvB⇒v=E/BqE = qvB \Rightarrow v = E/B. The radius of the circular path in the magnetic field is r=mvqB=mEqB2r = \frac{m v}{q B} = \frac{m E}{q B^2}. q/m=ErB2=2000.005×0.42=250000 C/kgq/m = \frac{E}{r B^2} = \frac{200}{0.005 \times 0.4^2} = 250000 \text{ C/kg}.

All Questions in this Chapter

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