Question

Calculate the radius of the orbit of a proton moving at 1.7x10^6 m/s in a magnetic field 0.8 T where v and B are perpendicular. Give your answer in centimetres.

Answer 1

2.28 cm

Step-by-step explanation:

Radius of the proton if v and B is perpendicular will be,

`r=(mv)/(qB)`

Now,

mass of proton,
`m=1.67*10^(-27) kg`.

Radius of the orbit,
`r=1.7*10^(6) m/s`.

Magnetic field,
`B=0.8 T`.

Charge on the proton,
`q=1.6*10^(-19) C`.

Now substitute all the variables in r.

`r=(1.67*10^(-27) kg(1.7*10^(6) m/s)  )/(0.8 T(1.6*10^(-19) C) )`.

`r=2.28 *10^(-2) m\\r=2.28 cm`.

Therefore the radius of the orbit of moving proton is 2.28 cm.