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 - General Problems - Page 588: 85

Answer

a) Negative. b) $ p=q B_0\left[\dfrac{L^2+d^2}{2d}\right] $

Work Step by Step

a) According to the right-hand rule and since the particle moves up, the particle is negatively charged. Therefore, $q$ is negative. b) We know that the momentum of a particle is given by $$p=mv\tag 1$$ We also know that the particle was moving horizontally, then entered a magnetic field that exerts a force on it. We also know that the magnetic field forces the particle to move on a circular path (for a small distance which is an arc) which means that the net force exerted on the electron is given by $$\sum F=F_B=\dfrac{mv^2}{r}\tag 2$$ whereas $r$ is the radius of curvature, as you see in the figure below. We know that the magnetic force is given by $$F_B=qvB_0$$ Plugging into (2); $$ qvB_0=\dfrac{mv^2}{r} $$ $$ q B_0=\dfrac{\overbrace{mv}^{p} }{r} $$ Hence, $$p=q B_0r\tag 3$$ Now we need to find $r$ in terms of $d$ and $L$. From the geometry of the figure below, and by applying Pythagorean therom $$r^2=L^2+(r-d)^2=L^2+r^2-2rd+d^2$$ Thus, $$0=L^2- 2rd+d^2$$ Solving for $r$; $$r=\dfrac{L^2+d^2}{2d}$$ Plugging into (3); $$\boxed{p=q B_0\left[\dfrac{L^2+d^2}{2d}\right]}$$
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