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QPages 230-234: Q32: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q32 Solution: An electron of kinetic energy 100 eV circulates in a path of…

An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

Step-by-Step Verified Solution

**Answer:** B=3.37×10−4 TB = 3.37 \times 10^{-4} \text{ T}, f=5.36×107 rev/sf = 5.36 \times 10^7 \text{ rev/s}. **Step-by-Step Explanation:** KE=100 eV=1.6×10−17 JKE = 100 \text{ eV} = 1.6 \times 10^{-17} \text{ J}. v=2KE/m=5.93×106 m/sv = \sqrt{2KE/m} = 5.93 \times 10^6 \text{ m/s}. B=mver=9.1×10−31×5.93×1061.6×10−19×0.1=3.37×10−4 TB = \frac{m v}{e r} = \frac{9.1 \times 10^{-31} \times 5.93 \times 10^6}{1.6 \times 10^{-19} \times 0.1} = 3.37 \times 10^{-4} \text{ T}. f=v2πr=9.44×106 Hzf = \frac{v}{2\pi r} = 9.44 \times 10^6 \text{ Hz}.

All Questions in this Chapter

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