Question
A uniform magnetic field points north; its magnitude is $1.5 \mathrm{T} .$ A proton with kinetic energy $8.0 \times 10^{-13} \mathrm{J}$ is moving vertically downward in this field. What is the magnetic force acting on it?
Step 1
We know that the kinetic energy of a particle is given by the equation $K = \frac{1}{2} m v^2$, where $m$ is the mass of the particle and $v$ is its speed. We can rearrange this equation to solve for $v$: \[v = \sqrt{\frac{2K}{m}}\] Show more…
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