Question

Singly charged positive ions are kept on a circular orbit in a cyclotron. The magnetic field inside the cyclotron is 1.833 T. The mass of the ions is 2.00×10-26 kg, and speed of the ions is 2.05 percent of the speed of the light. What is the diameter of the orbit? (The speed of the light is 3.00×108 m/s.)

Answer 1

Answer:

The diameter is  0.8376 m

Explanation:

Magnetic force is the force that is associated with the magnetic field, the magnitude of the magnitude force can be obtained using equation 1;

F = q v B  .....................................1,

where q is the magnitude of the charge of the particle =;

v is the  velocity and;

B is the magnetic field = 1.833 T ;

Here, the path of the charge is circular so the  force can be also considered as the centripetal force  which is represented in equation 2;

$F_{c}$ = m $v^{2}$ / r ....................................2,

m is the particle's mass = 2.00 x$10^{-26}$ kg

v is the speed of ion =  2.05% of 3.00 x $10^{8}$ = 0.0205 x 3.00 x $10^{8}$

                                    = 6.15 x  $10^{6}$ m/s ;

Singly charged ion has a charge equal to the electron charge and the magnitude = 1.60217646 ×$10^{-19}$ C and;

r is the radius of the circular path.

to get  the diameter of the orbit  we equate equation 1 to 2 and isolate r in equation 2.

q v B = m $v^{2}$ / r

r = m v/q B.....................................3

r = (2.00 x$10^{-26}$ kg) x (6.15 x  $10^{6}$ m/s)  / (1.60217646 ×$10^{-19}$  C) x (1.833 T)

r = 0.4188 m

The diameter is r x 2

D = 0.4188 m x 2

D = 0.8376 m

Therefore the diameter is  0.8376