Discussion
There is a space region where \(\overline{E} = E_{0}\hat{z}\) and \(\overline{B} = B_{0}\hat{x}\) are both present. A charged particle with velocity \(\left(\frac{E_{0}}{B_{0}}\right)\hat{y}\) enters this region. The trajectory of the particle in this region is a :
Answer: A
The trajectory is a Straight line.
The Lorentz force on the particle is \(q(\overline{E} + \overline{v} \times \overline{B})\). With \(\overline{v} = (E_{0}/B_{0})\hat{y}\) and \(\overline{B} = B_{0}\hat{x}\):
\[ \begin{aligned} \overline{v} \times \overline{B} &= \frac{E_{0}}{B_{0}}B_{0}(\hat{y} \times \hat{x}) \\ &= -E_{0}\hat{z} \end{aligned} \]
This exactly cancels \(\overline{E} = E_{0}\hat{z}\), so the net force is zero and the particle moves undeflected in a straight line. This is the principle of the velocity selector. A particle with any other speed would follow a cycloid-type path; circles and helices need a magnetic field alone.
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