Question

1 Mass Measurement What we do to measure the mass of subatomic particles is to shoot them into a magnetic field and measure the resulting circular path. We set up the magnetic field region with a current in two Helmholz coils (see your homework problem 30.61). Basically, this means that the field has equal contributions from each coil; each coil produces a magnetic field measured some distance z away from the center of N loops of wire: Nuo R B= 2(R2 + z23372 Here, we have 2 coils (R = 15.5 cm) whose B fields add together midway between them (z = 7.75 cm from each coil), and we have N = 130 turns for each coil. 1. Bu If the current is 2 A in each coil, calculate the total magnetic field due to both. 2. v= The electrons are introduced into the magnetic field region by accelerating them through a voltage of 300 V. By conserving energy, write down an equation for the speed of the elec- trons as a function of their mass 3. me Ok, remember we said the radius of the circular path would be r= mv eB for electrons. When doing this experiment, you observe that the r = 4 cm. So if you substitute your expression for v, you should be able to solve for the mass of the electron! Look up the exact (accepted) value for me and just calculate the percentage differ- ence between what we got and the “real” value. Not bad for measuring tiny masses on a tabletop! 4.

Answer 1

NHOR? I B 2(R722) 3/2 Given, N=130; R=15.5 = 0.155 m. Is lA. cm Z = 7 -75 uns 0.0775 m. T. is B 130 * 4*x 10-7x (0.15592x2 [10'155)2 + (0.0775)2] ชา2 = 7.54*10-4 Tesla Kif en al2 mo2 Given, v=8oov, X 300 12 eV 2*1.6x10-19 ve = . 2 me me 2 et ma me Da me v G C = e B eB 2me e v amer to 19 e e B 2 ra e (7.54x1024) 37 (0.04) B 16x10-19 2 x 300 me = 28 kg ime=2.4266x10-31 A Axact value, me = 9.11*10-31 percentage diffo kg. (me) exact me X 100 (he) exact a 73. 3 60%