m edugen.wileyplus.com Wiley PLUS Halliday, Fundamentals of Physics, 10e Assignm

Question

m edugen.wileyplus.com Wiley PLUS Halliday, Fundamentals of Physics, 10e Assignment open Assignment Chapter 28, Problem 017 An alpha particle (q +2e, m 4.00 u) travels in a circular path of radius 5.75 cm in a uniform magnetic field with B 1.75 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy. units (a) Number units (b) Number (C Question Attempts o of 8 used SAM ISEMITANSwERI D 2000-2017 by John Wily &Sons;, Inc or nelated companies. Al rights reserved

Explanation / Answer

(a) Here the alpha particle is traveling in a circle.

Therefore, we apply the equation -

F = mv^2/r

Now -
F (magnetic field) = qvXB = qvB
Set the two equal, mv^2/r = qvB
v = rqB/m

Now put the values -

v = (0.0575)(2*1.6e-19)(1.75)/(4*1.66e-27) = 2.77 x 10^6 m/s

(b) Period = 2pi*r/v = 2(3.14)(0.0575)/(2.77x10^6) = 1.30 x 10^-7 s

(c) Here, v << c, So, we don't need to worry about relativity.

Therefore,
KE = 0.5*m*v^2 = 0.5(4*1.66e-27)(2.77e6)^2 = 2.547 x 10^-14 J = (2.547x10^-14) / (1.6x10^-19)

= 1.592 x 10^5 eV = 159.2 keV

(d) Assuming all the potential energy went to kinetic energy, PE = qV = KE

V = (2.547 x 10^-14)/(2*1.6e-19) = 7.96 x 10^4 V = 79.6 kV

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