Sample Problem Uniform circular motion of a charged particle in a magnetic field

Question

Sample Problem Uniform circular motion of a charged particle in a magnetic field Figure 28-12 shows the essentials of a mass spectrometer, which can be used to measure the mass of an ion; an ion of mass m (to be measured) and charge q is produced in source S. The initially stationary ion is accelerated by the electric field due to a potential difference V.The ion leaves S and enters a separator chamber in which a uniform mag- netic field B is perpendicular to the path of the ion. A wide detector lines the bottom wall of the chamber, and the B causes the ion to move in a semicircle and thus strike the detector. Suppose that B = 80.000 mT, V = 1000.0 V, and ions of charge q = +1.6022 × 10-19 C strike the detector at a point that lies at x = 1.6254 m. What is the mass m of the individual ions, in atomic mass units (Eq. 1-7: 1 u = 1.6605 × 10-27 kg)? Detrctor +g Fig. 28-12 Essentials of a mass spectrometer. A positive ion, after being accelerated from its source S by a potential dif- KEY IDEASference V,enters a chamber of uniform magnetic field B There it travels through a semicircle of radius r and strikes a detector at a distance x from where it entered the chamber (1) Because the (uniform) magnetic field causes the (charged) ion to follow a circular path, we can relate the ion's mass m to the path's radius r with Eq. 28-16 (r= mv/glB). From Fig. 28-12 we see that r = x/2 (the radius is half the di- ameter). From the problem statement, we know the magni- tude B of the magnetic field. However, we lack the ion's speed v in the magnetic field after the ion has been acceler ated due to the potential difference V. (2) To relate v and V, we use the fact that mechanical energy (EeK+U)is conserved during the acceleration. or (28-22) 71 Finding mass: Substituting this value for v into Eq. 28-16 gives Us Finding speed: When the ion emerges from the source, its kinetic energy is approximately zero. At the end of the acceleration, its kinetic energy is mv2. Also, during the ac- celeration, the positive ion moves through a change in potential of -V.Thus, because the ion has positive charge q its potential energy changes by -qV. If we now write the conservation of mechanical energy as Thus, Solving this for m and substituting the given data yield B'qr2 8V 0.080000 T 16022 x 10-19 C) (1.6254 m) we get 8(1000.0 V) = 3.3863 × 10-25 kg = 203.93 u. (Answer) Additional examples, video, and practice avallable at WileyPLUS

Explanation / Answer

1) .0.5mv2=qV

v=?(2qV/m)

2). r = mv/Bq

x=2r

=> displacement= 2mv/Bq = 2/B * ?(2mV/q)

3). distance = ?*r

time = distance /speed

time= (?*mv/Bq) / v

=> time = ?*m/Bq

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