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chp29_explained_000

# chp29_explained_000 - Chapter 29 Question 13 A region...

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Chapter 29 Question 13 A region contaings an electric field E = 7.4 i ˆ + 2.8 j ˆ kN / C and a magnetic field B =15 j ˆ +36 k ˆ mT. Find the electromagnetic force on (a) a stationary proton, (b) an electron moving with velocity v = 6.1 i ˆ Mm / s. (a) We’ll use the Lorentz force law, F =q( E + v × B ). For part (a), v = 0. N F or N j i F N j i F E q F 15 16 16 3 19 10 27 . 1 ˆ 10 49 . 4 ˆ 10 9 . 11 10 ) ˆ 8 . 2 ˆ 4 . 7 ( 10 602 . 1 ! ! ! ! " = " + " = " + # " = = v v v v (b) Next we use the same equation with the given velocity. Since we’re doing an electron this time, we’ll use negative e. ( ) ( ) ( ) ( ) ( ) ( ) N F or N k j i F N j k j i F N k j i j i F B v E q F 15 16 3 19 3 6 3 19 10 7 . 37 10 ˆ 6 . 146 ˆ 3 . 347 ˆ 9 . 11 10 ˆ 6 . 219 ˆ 5 . 91 ˆ 8 . 2 ˆ 4 . 7 10 602 . 1 10 ˆ 36 ˆ 15 ˆ 10 1 . 6 10 ˆ 8 . 2 ˆ 4 . 7 10 602 . 1 ! ! ! ! ! " = " ! + ! = " ! + + " ! = " + " " + " + " ! = " + = v v v v v v v Question 22 Show that the orbital radius of a charged particle moving at right angles to a magnetic field B can be written qB Km r 2 = where K is the kinetic energy in joules, m the particle mass, and q its charge. To solve this problem you can either use equation 29-3 and relate v to K (K= ½ mv 2 ), or you can derive the equation from first principles. In 2A you learned that if an object moves in a circle, it must be experiencing a force F=mv 2 /r to keep it moving in that circle. Then we observe that for a particle moving perpendicular to the B field, F=qvB. Then we can equate the two forces.

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qB K r qB m K m r m K v mv K qB mv r qvB r mv 2 2 2 2 1 2 2 = = = = = = Problem 27
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chp29_explained_000 - Chapter 29 Question 13 A region...

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