003 (part 1 of 4) 10.0 points In a linear accelerator, protons are acceler- ated

Question

003 (part 1 of 4) 10.0 points In a linear accelerator, protons are acceler- ated from rest through a potential difference to a speed of approximately 3.77 × 10° m/s The resulting proton beam produces a current of 1.4 × 10-6 A 27 The mass of the proton is 1.67 x 10 kg -19 C and its charge on the proton is 1.6 × 10-19 C Determine the potential difference through which the protons were accelerated. Answer in units ofV 004 (part 2 of 4) 10.0 points If the beam is stopped in a target, deter mine the amount of thermal energy that is produced in the target in one minute. Answer in units of J. 005 (part 3 of 4) 10.0 points The proton beam enters a region of uniform magnetic field B, as shown below, that causes the beam to follow a semicircular path.

Explanation / Answer

In linear accelerators particle will acquire energy due to potential difference only ,
Given

   proton mass m = 1.67*10^-27 kg, charge q = 1.6*10^-29 C, speed v= 3.77*10^6 m/s,i = 1.4*10^-6 A

here energy transfered E = Q*V , kinetic enrgy => 1/2m v^2 = Q*V ==> V = (0.5* mv^2/Q)

003.    potential difference is V = (0.5*1.6*10^-27*(3.77*10^6)^2)/(1.6*10^-19)
              = 71064.5 V

004
          
here per one sec there are 3.77*10^6 protons moving then the kinetic energy associated with them is

   k.e = 1/2 m v^2 = 0.5*1.67*10^-27*(3.77*10^6)^2
           = 1.18677715*10^-14 J,per proton

   k.e per min = 1.18677715*10^-14*60*3.77*10^6 = 2.6844899133*10^-6 J per min

005.

  
cyclotron, centripetal force = magnetic force acting on the particle
       mv^2 /R = Bqv Sin theta theta is 90 in circular path

       B= mV/qR = (1.67*10^-27*3.77*10^6)/(1.6*10^-19*0.101)
= 0.3895977722772 T

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