Question

Protons with kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field, so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.

Answer 1

Given:
Kinetic energy of proton = K
Distance of the target from the accelerator = l
Therefore, radius of the circular orbit ≤ l
As per the question, the beam is bent by a perpendicular magnetic field.
We know   
`r = (mv)/(eB)`
For a proton, the above equation can be written as:
`l  = (m_pv)/(eB)`      (As r=l)....(i)
Here,
m_p is the mass of a proton
 v is the velocity
 e is the charge
 B is the magnetic field
`1/2 m_pv^2 = K`

⇒ `v= sqrt((2K)/m_p`

putting the value of V in the equation (i),we get
`l =( sqrt2K_mp)/(eB)`

`B = sqrt(2Kmp)`