ooo U.S. Cellular LTE 10:39 PM 7490.. Question Done Problem 20.19 A beam of prot

Question

ooo U.S. Cellular LTE 10:39 PM 7490.. Question Done Problem 20.19 A beam of protons is accelerated through a potential difference of 0.750kV and then enters a uniform magnetic field traveling perpendicular to the field. Part A What magnitude of field is needed to bend these protons in a circular arc of diameter 1.71m 1 2 3 4 5 678 9 0 Submit Give Up Incorrect; Try Again; 4 attempts remaining Part B What magnetic field would be needed to produce a path with the same diameter i the particles were electrons having the same speed as the protons

Explanation / Answer

A)

kinetic enrgy gained by proton, KE = q*delta V

0.5*m*v^2 = q*delta V

v = sqrt(2*q*delta V/m)

= sqrt(2*1.6*10^-19*750/(1.67*10^-27) )

= 3.79*10^5 m/s

now, radius of the orbit, r = d/2 = 1.71/2 = 0.855 m

now Apply, q*v*B = m*v^2/r

==> B = m*v/(r*q)

= 1.67*10^-27*3.79*10^5/(0.855*1.6*10^-19)

= 4.63*10^-3 T

B) for elelctron

kinetic enrgy gained by proton, KE = q*delta V

0.5*m*v^2 = q*delta V

v = sqrt(2*q*delta V/m)

= sqrt(2*1.6*10^-19*750/(9.1*10^-31) )

= 1.624*10^7 m/s

now, radius of the orbit, r = d/2 = 1.71/2 = 0.855 m

now Apply, q*v*B = m*v^2/r

==> B = m*v/(r*q)

= 9.1*10^-31*1.624*10^7/(0.855*1.6*10^-19)

= 1.08*10^-4T

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