Question
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of $1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}$ undergoes an acceleration of $4.0 \times 10^{16} \mathrm{~m} / \mathrm{s}^{2}$ to the right (the positive $x$ -direction) when its velocity is upward (the positive $y$ -direction). Determine the magnitude and direction of the field.
Step 1
In this case, the electron is moving perpendicular to the magnetic field, so $\theta = 90^{\circ}$ and $sin\theta = 1$. Show more…
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