hw6.5 - x 1 ' at which deuterons exit the magnetic field...

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HW 6.5 Given: F i n d : C q p 19 10 60 . 1 ? a V kg m p 27 10 67 . 1 ? 1 x ? 1 s m v / 10 1 . 3 6 0 x T B 3 . 2 p D q q p D m m 2 Protons ( q p = 1.60 × 10 -19 C, m p = 1.67 × 10 -27 kg) are accelerated from rest through a potential difference V a and emerge moving in the + y direction with a speed v 0 = 3.1 × 10 6 m/s. After passing through a field-free region, they enter at the origin a region ( y > 0) with a uniform magnetic field B = 2.3 T in the + z direction (out of the page). They then execute a semicircular trajectory and emerge from the magnetic field region at distance x 1 from the origin. (a) What is the magnitude of the potential difference V a ? (b) Calculate the value of x 1 at which the protons exit the magnetic field region. (c) Suppose instead deuterons ( q D = q p , m D = 2 m p ) are accelerated from rest through the same potential V a and enter the same magnetic field. Calculate the position
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Unformatted text preview: x 1 ' at which deuterons exit the magnetic field region. Solutions a.) To find a V we can use the following formula: KE U 2 2 1 v m V q p a p p p a q v m V 2 2 b.) To find we can use the following formula: 1 x B q v m R p p B q v m R x p p 1 2 2 B q v m x p p 1 2 c.) To find we can use the following formula: 1 x D D a q v m V 2 2 D p D p D D p p D D a m q q v m m q q v m m q V v 2 2 2 2 2 D p p D D p p D D p p D D D p D p D D D p D p D D D q q m m B v B q q v m m B q q v m m B q m q q v m m B q m q q v m m B q v m R 2 2 2 2 2 2 2 2 D p p D q q m m B v R x 1 2 2 D p p D q q m m B v x 1 2...
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This note was uploaded on 01/26/2012 for the course PHYSICS 211 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.

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hw6.5 - x 1 ' at which deuterons exit the magnetic field...

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