Question

A particle with charge q and kinetic energy KE travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for its speed v and its mass m. (Use any variable or symbol stated above as necessary.) (a) expression for its speed (Do not use m in your answer.) v = (b) expression for its mass (Do not use v in your answer.) m =

Answer 1

a)
`v=(2.KE)/(qBR)`

b)
`m=((qBR)^2)/(2.KE)`

Step-by-step explanation:

Given that

Charge = q

Magnetic filed = B

Radius = R

We know that kinetic energy KE

`KE=(1)/(2)mv^2` ----------1

m v² = 2 .KE

The magnetic force F = q v B

Radial force

`Fr=(1)/(R)mv^2`

For uniform force these two forces should be equal

`q v B=(1)/(r)mv^2`

q v B R =m v²

q v B R = 2 .KE

`v=(2.KE)/(qBR)`

Now put the velocity v in the equation

`KE=(1)/(2)mv^2`

`m=(2 .KE)/(v^2)`

`m=(2.KE)/(\left((2.KE)/(qBR)\right)^2)`

`m=((qBR)^2)/(2.KE)`