Question

In a cyclotron final kinetic energy of a particle revolving in a circular path of radius ‘R’ exit is ______.

पर्याय

• K.E. ∝ R_exit

• K.E. ∝ R²_exit

• `K.E. ∝ 1/(R_"exit")`

• `K.E. ∝ 1/(R_"exit"^2)`

Answer 1

In a cyclotron final kinetic energy of a particle revolving in a circular path of radius ‘R’ exit is `bbunderline(K.E. ∝ R_"exit"^2)`.

Explanation:

The magnetic force provides the centripetal force required for circular motion:

`qvB = (mv^2)/R`

v = `(qBR)/m`

Plug the velocity into the standard kinetic energy equation:

`K.E. = 1/2 mv^2`

= `1/2 m ((qBR)/m)^2` 

= `1/2 m((q^2B^2R^2)/(m^2))` 

= `(q^2B^2R^2)/(2m)` 

Since q, B, and m are constants for a specific particle and cyclotron setup, we see that:

K.E. ∝ R²