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Question Number 13965 by tawa tawa last updated on 25/May/17

$$\mathrm{A}\mathrm{cathode}\mathrm{ray}\mathrm{beam}\mathrm{is}\mathrm{bent}\mathrm{in}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}2\mathrm{cm}\mathrm{by}\mathrm{uniform}\mathrm{field}\mathrm{with}$$$$\mathrm{B}=4.5\times {10}^{-3}\mathrm{T}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{electrons}??.$$

Answered by sandy_suhendra last updated on 26/May/17

$$\mathrm{mass}\mathrm{of}\mathrm{electron}=\mathrm{m}=9.1\times {10}^{-31}\mathrm{kg}$$$$\mathrm{charge}\mathrm{of}\mathrm{electron}=\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$$$$\mathrm{B}=4.5\times {10}^{-3}\mathrm{T}$$$$\mathrm{R}=2\mathrm{cm}=0.02\mathrm{m}$$$${\mathrm{F}}_{\mathrm{Lorentz}}={\mathrm{F}}_{\mathrm{sf}}$$$$\mathrm{B}.\mathrm{e}.\mathrm{v}=\frac{\mathrm{m}.{\mathrm{v}}^2}{\mathrm{R}}$$$$\mathrm{v}=\frac{\mathrm{B}.\mathrm{e}.\mathrm{R}}{\mathrm{m}}=\frac{4.5\times {10}^{-3}\times 1.6\times {10}^{-19}\times 0.02}{9.1\times {10}^{-31}}=1.58\times {10}^7\mathrm{m}/\mathrm{s}$$

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