A narrow beam of protons having kinetic energy K emerges from an...

The momentum of a proton with kinetic energy $K$ is given by $p = \sqrt{2 m_p K}$.

When the proton beam enters a perpendicular magnetic field $B$, it follows a circular path of radius $R$ given by:

$R = \dfrac{p}{eB} = \dfrac{\sqrt{2 m_p K}}{eB}$

The protons emerge from the accelerator and move towards the plane target. The maximum forward distance traveled by a particle moving in a circular path (starting from the origin and tangent to the initial velocity) is equal to its radius $R$.

For the beam to just miss the plane target kept at a distance $l$ in front of the accelerator, the maximum forward distance must be equal to $l$.

Therefore, $R = l$.

Substituting the value of $R$:

$l = \dfrac{\sqrt{2 m_p K}}{eB}$

Rearranging the above equation to solve for the magnetic field $B$:

$B = \dfrac{\sqrt{2 m_p K}}{e l}$