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.

#### Solution

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)`