P29_062 - (a Using Eq 29-3(with angle φ equal to 90 ◦ we...

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62. With the ~ B pointing “out of the page,” we evaluate the force (using the right-hand rule) at, say, the dot shown on the left edge of the particle’s path, where its velocity is down. If the particle were positively charged, then the force at the dot would be toward the left, which is at odds with the Fgure (showing it being bent towards the right). Therefore, the particle is negatively charged; it is an electron.
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Unformatted text preview: (a) Using Eq. 29-3 (with angle φ equal to 90 ◦ ), we obtain v = | ~ F | e | ~ B | = 4 . 99 × 10 6 m / s . (b) Using either Eq. 29-14 or Eq. 29-16, we Fnd r = 0 . 00710 m. (c) Using Eq. 29-17 (in either its Frst or last form) readily yields T = 8 . 93 × 10 − 9 s....
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