That is eq 27. 11 R= mv/(|q|B) for number 2 A beam of relativistic electrons tra

Question

That is eq 27. 11 R= mv/(|q|B) for number 2

A beam of relativistic electrons traveling in the +x direction enters a region of width L = 3.73 cm containing a uniform magnetic field oriented in the -z direction, as illustrated in the figure below. The speed of the electrons is 99.992% the speed of light, or v = 0.99992 c (that's four nines and a two). The relativistic factor gamma for the electrons is defined to be gamma= [1- v2/c2]/-1/2. What is gamma for these electrons? The formula for the radius of curvature R of a charged particle in a magnetic field, eq. (27.11) on p. 893 of the textbook, happens to apply to relativistic particles, as long as the mass factor m is replaced by gamma m. What magnetic field strength B is required to deflect the electrons by an angle of theta = 11.25 degree by the time they exit the magnetic field? (This is the deflection angle for the relativistic electron beam that drives the free-electron laser on the ground floor of Watanabe Hall.)

Explanation / Answer

a) relativistic factor=79.058. b) 8.912e-3 Tesla. from R=mv/qB and L/R=sin(theta).

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