To find the magnitude of the magnetic field, we can use either the proton or the electron.
The magnetic force experienced by a charged particle moving in a magnetic field is:
[tex]F=qvB \sin \theta[/tex]
where
q is the charge
v is the velocity of the particle
B is the magnitude of the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B.
Let's take the proton. The particle is moving in the x-direction, while the force is in the y-direction: for the right-hand rule, this means that the magnetic field B is in the z-direction, so v and B are perpendicular and [tex]\theta=90^{\circ}[/tex], so [tex]\sin \theta=1[/tex] and we can ignore it in the formula.
Therefore, we have:
[tex]F=qvB[/tex]
or
[tex]B= \frac{F}{qv} [/tex]
and since we know the force intensity, the velocity of the proton v and its charge ([tex]q=+1.6 \cdot 10^{-19}C[/tex]), we can find the magnitude of B:
[tex]B= \frac{2.10 \cdot 10^{-16}N}{(1.6 \cdot 10^{-19}C)(1200 m/s)}=1.09 T [/tex]
