Question

Barium ion with a 2+ charge travels at a speed 800 m/s and is acted upon by a magnetic force having a magnitude of 1700.0 N. Determine the magnitude of magnetic field acting perpendicularly on Ba^(2+).

Answer 1

$$F=qvB\sin\theta$$

$$B=\frac{F}{qv\sin\theta}\\ =\frac{1700}{2*1.6*10^{-19}*800*\sin90^\circ}=6.64*10^{18}\:\rm T$$