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

The diameter is 0.8376 m

Step-by-step 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