## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 19 - Problems - Page 760: 107

#### Answer

(a) $f = 1.2 \times 10^7~Hz$ (b) $f = 2.2 \times 10^{10}~Hz$

#### Work Step by Step

(a) $F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find an expression for the speed $v$: $\frac{mv^2}{r} = qvB$ $v = \frac{q~B~r}{m}$ We can find an expression for the frequency: $f = \frac{speed}{distance}$ $f = \frac{v}{2\pi~r}$ $f = \frac{(\frac{q~B~r}{m})}{2\pi~r}$ $f = \frac{q~B}{2\pi~m}$ $f = \frac{(1.6\times 10^{-19}~C)(0.80~T)}{(2\pi)~(1.67\times 10^{-27}~kg)}$ $f = 1.2 \times 10^7~Hz$ (b) $F = \frac{mv^2}{r}$ $F = qvB$ We can equate the two expressions for $F$ to find an expression for the speed $v$: $\frac{mv^2}{r} = qvB$ $v = \frac{q~B~r}{m}$ We can find an expression for the frequency: $f = \frac{speed}{distance}$ $f = \frac{v}{2\pi~r}$ $f = \frac{(\frac{q~B~r}{m})}{2\pi~r}$ $f = \frac{q~B}{2\pi~m}$ $f = \frac{(1.6\times 10^{-19}~C)(0.80~T)}{(2\pi)~(9.1\times 10^{-31}~kg)}$ $f = 2.2 \times 10^{10}~Hz$

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