A magnet produces a $0.30-\mathrm{T}$ field between its poles, directed to the east. A dust particle with charge $q=-8.0 \times 10^{-18} \mathrm{C}$ is moving straight down at $0.30 \mathrm{cm} / \mathrm{s}$ in this field. What is the magnitude and direction of the magnetic force on the dust particle?
Added by Zachary D.
Step 1
0 \times 10^{-18} \, \text{C} \), \( \vec{v} = -0.003 \, \text{m/s} \, \hat{k} \), and \( \vec{B} = 0.3 \, \text{T} \, \hat{i} \). Show more…
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