Question

Singly charged positive ions are kept on a circular orbit in a cyclotron. The magnetic field inside the cyclotron is 1.981 T. The mass of the ions is 3.67×10-26 kg, and speed of the ions is 2.75 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.)(in m)

Answer 1

Therefore the diameter of the orbit is 1.91m

Step-by-step explanation:

The path followed by the particle is a circle that keeps the relation given by

`BVq=(mv^2)/(r)`

Re-arrange the equation for r,

`r=(mv)/(Bq)`

From V is equal to speed of ions and there is a procent for this 'real speed' we have

`V=(2.75)/(100) 3*10^8m/s`

`V= 8.25*10^6m/s`

Replacing in the previous equation,

`r= ((3.67*10^(-26))(8.25*10^6))/((1.981)(1.6*10^(-19)))`

`r=0.955m`

`D=r*2= 1.91m`

Therefore the diameter of the orbit is 1.91m