The paper has talked about the strong magnetic field will trap the low energy proton. So, how is the field strength and the proton energy relationship?

the proton moving radius can be formulated by:

$R = \frac {m v}{e B}$

and according the special relativity

$v = c \frac{p c}{E} = c \sqrt { 1- \left( \frac {m c^2}{E} \right) ^2 }$

Thus,

$R = \frac { m c} { e B } \sqrt { 1- \left(\frac {m c^2}{E} \right)^2 }$

Sub all the constant

$R = 3.129738 \frac{1}{B} \sqrt { 1- \left(\frac {938 MeV }{E} \right)^2 }[m][T]$

Thus, for slow proton, say 50MeV, the radius is

$R = 983.02 / B [mm][T]$

even for 1MeV

$R = 144.40 / B [mm][T]$

but for 1keV

$R = 4.57 /B [mm][T]$