Problem 74-2
Source: Problem developed by the site author.
A proton with a kinetic energy equal to 7 MeV is launched horizontally into a laboratory
chamber that has a magnetic induction field equal to 2.80 T . This field is oriented
vertically downwards and the proton moves from left to right, as shown in Figure 74-02.1 .
Considering the mass of the proton equal to 1.67 x 10-27 Kg, determine:
a) The force that the proton experiences when entering the chamber.
b) Calculate the radius of the circumference that the proton describes.
c) Calculate the proton angular frequency.
Figure 74-02.2
Solution of the Problem 74-2
Item a
As was studied in the theoretical part, item 2.1, when a particle with an electric charge
penetrates a magnetic induction field, it undergoes a lateral force and, as a consequence,
it starts to perform a circular motion. To determine the force to which the proton is subjected
we must use eq. 74-02. For clarity we will reproduce the eq. 74-02, below:
eq. 74-02
By the eq. 74-02 we don't know the value of the proton's speed. However, as it was provided
the kinetic energy of it and its mass, we can calculate. So, based on the energy equation
kinetics, K, and after an algebraic arrangement, we have:
v = (2 K/m)1/2
Substituting for numeric values we have:
v = (2 x 7 MeV x 1.60 X 10-13 J/MeV/ 1.67 x 10-27 Kg)1/2
Carrying out the calculation, we find:
v = 3.66 x 107 m/s
By the data of the problem the direction of the proton and the magnetic induction field are
orthogonal to each other. So θ = 90° and sen 90° = 1.
And the charge of the proton is worth q = 1.6 x 10 -19 C .
Now, as we have all the necessary data, we can substitute the numerical values in eq. 74-02 , or:
Fm = 1.6 x 10-19 x 3.66 x 107 x 2.80 x 1
Performing the calculation we find the value of the magnetic force acting on the proton, or
Fm = 1.64 x 10-11 N
Item b
For the determination of the radius of the circumference we will use
eq. 74-05, reproduced below for clarity.
eq. 74-05
Thus, replacing with the numerical values we have:
r = 1.67 x 10-27 x 3.66 x 107 / 1.60 x 10-19 x 2.80
Performing the calculation, we find
r = 0.136 m = 13.6 cm
Item c
To find the value of the angular frequency we will use eq. 74-07, reproduced below.
eq. 74-07
Therefore, making the substitution by numeric values we have:
