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QPages 230-234: Q30: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q30 Solution: A particle of charge and mass is projected with a speed of in a …

A particle of charge 2.0×10−8 C2.0 \times 10^{-8} \text{ C} and mass 2.0×10−10 g2.0 \times 10^{-10} \text{ g} is projected with a speed of 2.0×103 m s−12.0 \times 10^3 \text{ m s}^{-1} in a region with a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.

Step-by-Step Verified Solution

**Answer:** r=0.125 mr = 0.125 \text{ m}, T=3.9×10−4 sT = 3.9 \times 10^{-4} \text{ s}. **Step-by-Step Explanation:** q=2×10−8 Cq = 2 \times 10^{-8} \text{ C}, m=2×10−13 kgm = 2 \times 10^{-13} \text{ kg}, v=2×103 m/sv = 2 \times 10^3 \text{ m/s}, B=0.1 TB = 0.1 \text{ T}. r=mvqB=2×10−13×2×1032×10−8×0.1=0.2 mr = \frac{m v}{q B} = \frac{2 \times 10^{-13} \times 2 \times 10^3}{2 \times 10^{-8} \times 0.1} = 0.2 \text{ m}. T=2πmqB=2π×2×10−132×10−8×0.1=6.28×10−4 sT = \frac{2\pi m}{q B} = \frac{2\pi \times 2 \times 10^{-13}}{2 \times 10^{-8} \times 0.1} = 6.28 \times 10^{-4} \text{ s}.

All Questions in this Chapter

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