In a linear accelerator, protons are accelerated from rest through a potential d

Question

In a linear accelerator, protons are accelerated from rest through a potential difference to a speed of approximately 3.1 x 106 meters per second. The resulting proton beam produces a current of 2 x 10-6 ampere. Determine the potential difference through which the protons were accelerated. If the beam is stopped in a target, determine the amount of thermal energy that is produced in the target in one minute. The proton beam enters a region of uniform magnetic field B. as shown above, that causes the beam to follow a semicircular path. Determine the magnitude of the field that is required to cause an are of radius 0.10 meter. What is the direction of the magnetic field relative to the axes shown above on the right?

Explanation / Answer

Given that speed of the proton is v = 3.1*10^6 m/s Proton beam produces a current i = 2*10^-6 A a ) Potential difference   q V = 1/2 mv^2          1.6*10^-19 C * V = 1/2 * 1.67*10^-27 kg * ( 3.1*10^6 m/s)^2             V = 5.01*10^4 V b ) Thermal energy produced in one mim time is       Power P = V * i                        = 5.01*10^4 V * 2*10^-6 A                        = 0.1 W        Energy E = P * t                         = 0.1 * 60s                          = 6.01 J c ) Given that radius of the semi circular path   r = 0.10 m magnetic field B = m v / q r                          = 1.67*10^-27 kg * 3.1*10^6 m/s / 1.6*10^-19 C * 0.10 m                         = 0.32 T d ) direction of the magnetic field is into the page

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