Question

A proton enters a magnetic field (B = 5.00 x 10^-2 T INTO THE PAGE) with

a velocity of 5.60 x 10^6 m/s West. Calculate the RADIUS (in m) of the
circular path the particle will follow when it is travelling within the
magnetic field. Will it move CLOCKWISE or COUNTERCLOCKWISE? Which
HAND RULE did you use?*

Answer 1

r = 1.16 m

Step-by-step explanation:

It is given that,

Magnetic field,
`B=5.6* 10^6\ m/s`

Velocity of proton,
`v=5.6* 10^6\ m/s`

We need to find the radius (in m) of the circular path the particle will follow when it is travelling within the magnetic field. The radius followed by the proton is given by :

`r=(mv)/(qB)`

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

So,

`r=(1.67* 10^(-27)* 5.6* 10^6)/(1.6* 10^(-19)* 5* 10^(-2))\\\\r=1.16\ m`

So, the radius of circular path is 1.16 m.