Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 20 - Magnetism - Problems - Page 584: 20

Answer

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Work Step by Step

Find the speed of each charged particle using energy conservation. The electrical potential energy becomes kinetic energy. $$qV=\frac{1}{2}mv^2$$ $$v=\sqrt{\frac{2qV}{m}}$$ When the particle enters the magnetic field, the magnetic force causes it to move in a circular path. Use the relationship from Example 20-6, page 567, and combine it with the above expression. $$r=\frac{mv}{qB} =\frac{1}{B}\sqrt{\frac{2mV}{q}}$$ We are asked to express the radius of the path for the deuteron in terms of that for the proton. $$\frac{r_d}{r_p}=\frac{\frac{1}{B}\sqrt{\frac{2m_dV}{q_d}}}{\frac{1}{B}\sqrt{\frac{2m_pV}{q_p}}}$$ $$\frac{r_d}{r_p}=\frac{\sqrt{\frac{m_d}{m_p}}}{\sqrt{\frac{q_d}{q_p}}}=\sqrt{2}$$ $$r_d=\sqrt{2}r_p$$ We are asked to express the radius of the path for the alpha particle in terms of that for the proton. $$\frac{r_\alpha}{r_p}=\frac{\frac{1}{B}\sqrt{\frac{2m_\alpha V}{q_\alpha }}}{\frac{1}{B}\sqrt{\frac{2m_pV}{q_p}}}$$ $$\frac{r_\alpha }{r_p}=\frac{\sqrt{\frac{m_\alpha }{m_p}}}{\sqrt{\frac{q_\alpha }{q_p}}}=\sqrt{2}$$ $$r_\alpha =\sqrt{2}r_p$$
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