What is the force on a \( 1.38 \mathrm{C} \) charge moving perpendicular to a \( 6.12 \mathrm{~T} \) magnetic field at \( 7.82 \mathrm{~m} / \mathrm{s} \) ?
\( 1.76 \mathrm{~N} \)
\( 34.7 \mathrm{~N} \)
\( 0.0289 \mathrm{~N} \)
\( 66.0 \mathrm{~N} \)

Answer

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Answer

\( F = 66.014 \mathrm{N} \)

Steps

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 ：\( F = 66.014 \mathrm{N} \)