Problem
What is the force on a \( 1.38 \mathrm{C} \) charge moving perpendicular to a \( 6.12 \mathrm{~T} \) magnetic field at \( 7.82 \mathrm{~m} / \mathrm{s} \) ? \( 1.76 \mathrm{~N} \) \( 34.7 \mathrm{~N} \) \( 0.0289 \mathrm{~N} \) \( 66.0 \mathrm{~N} \)
Solution
Step 1 ：\( F = qvB \sin \theta \)
Step 2 ：\( F = (1.38 \mathrm{C})(7.82 \mathrm{\frac{m}{s}})(6.12 \mathrm{T}) \sin \left(\frac{\pi}{2}\right) \)
Step 3 ：\( F = 66.014 \mathrm{N} \)
