The magnetic force on a moving charge is given by:
[tex] F=qvB \sin \theta [/tex]
where q is the charge, v is the speed of the charge, B is the magnetic field intensity and [tex] \theta [/tex] is the angle between the directions of B and v.
In our problem, the charge is an electron ([tex] q=1.6 \cdot 10^{-19}C [/tex]), the velocity of the charge is [tex] v=4.5 \cdot 10^4 m/s [/tex], the magnetic force is [tex] F=7.2 \cdot 10^{-18} N [/tex] and the angle between the direction of v and B is [tex] \theta=90^{\circ} [/tex], so [tex] \sin \theta=1 [/tex] and we can ignore the sine in the formula. Therefore, if we rearrange the equation and we put these numbers in, we can find the intensity of the magnetic field:
[tex] B=\frac{F}{qv}=\frac{7.2 \cdot 10^{-18} N}{(1.6 \cdot 10^{-19}C)(4.5 \cdot 10^{-4} m/s)}=0.001 T [/tex]
