# Repulsion force of a magnetic field and a positive ion

I'm trying to figure this out but my understanding is limited so I'd appreciate any help.

For the sake of simplicity, let's say I've stripped away the electron from an atom of hydrogen leaving only the proton. The charge of a proton is $1.6 × 10^{-19}$ C. According to Coulomb's Law, $$F= k\frac{qQ}{r^2},$$ So $k = 8.99 × 10^ 9$. Let's say $r = 0.5m$ .

Now, we have a magnetic field with a flux density (B) of $1.5$ T with positive perpendicular to the positively ionized hydrogen atom.

How can I figure out the net force and how much force the ion exerts on the magnet and the magnet exerts on the ion?

• If the charge is not moving, it is not influenced by the magnetic field. You're also missing most units. – Jasper May 24 '18 at 16:45
• For a magnetic force to work on the proton , you need it to be moving ? – Nehal Samee May 24 '18 at 16:45

You instead need the other half of the Lorentz Force Law: $\vec{F}_{magnetic} = q\vec{v} \times \vec{B}$, where $v$ is the velocity of the particle, and $B$ is the strength of magnetic field. Both are vectors, so you need the cross product.
You assumed a distance $r$ in your question, but it is unclear what that distance represents.