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Question Number 13872 by tawa tawa last updated on 24/May/17

$$\mathrm{An}\mathrm{alpha}\mathrm{particle}\mathrm{of}\mathrm{mass}6.68\times {10}^{-27}\mathrm{kg}\mathrm{and}\mathrm{charge}\mathrm{q}=+2,\mathrm{are}$$$$\mathrm{accelerated}\mathrm{from}\mathrm{rest}\mathrm{through}\mathrm{the}\mathrm{potential}\mathrm{difference}\mathrm{of}1\mathrm{kV}.\mathrm{it}\mathrm{then}\mathrm{enters}$$$$\mathrm{a}\mathrm{magnetic}\mathrm{field}\mathrm{B}=0.2\mathrm{T}\mathrm{perpendicular}\mathrm{to}\mathrm{their}\mathrm{direction}\mathrm{of}\mathrm{motion}.$$$$\mathrm{Calculate}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{their}\mathrm{path}.$$

Answered by ajfour last updated on 24/May/17

$$qV=\frac{1}{2}{mv}^2;alphaparticlegets$$$$acceleratedtospeed\mathit{v}$$$$\Rightarrow mv=\sqrt{2mqV}$$$$whenitentersregionofmagnetic$$$$fieldB,forceitexperiencesis$$$$F_B=qvB\sin \theta ;$$$$=qvB(\theta =\frac{\pi }{2}as\overline{B}is\mathrm{\perp }to\overline{\mathit{v}})$$$$insuchacasethemotioniscircular$$$$F_B=qvB=\frac{{mv}^2}{r}(Force=mass\times acc.)$$$$r=\frac{mv}{qB}=\frac{\sqrt{2mqV}}{qB}=\sqrt{\frac{2mV}{{qB}^2}}$$$$=\sqrt{\frac{2\times 6.68\times {10}^{-27}\times 1000}{2\times 1.6\times {10}^{-19}\times 4\times {10}^{-2}}}m$$$$=\sqrt{\frac{6.68}{6.4}\times {10}^{-3}}m=\sqrt{\frac{668}{64}}\times {10}^{-2}m$$$$=\sqrt{10+\frac{7}{16}}cm$$$$=\sqrt{10.438}cm\approx 3.23cm.$$

Commented by tawa tawa last updated on 24/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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