Question:

In the figure, the electric field \(E\) and the magnetic field \(B\) point to \(z\) and \(x\) directions, respectively, and have constant magnitudes. A positive charge \(q\) is released from rest at the origin. Which of the following statement(s) is/are true? \begin{center} \includegraphics[width=0.45\textwidth]{20.jpeg} \end{center}

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For charged particles in crossed \(\vec{E}\) and \(\vec{B}\) fields: \(\vec{E}\) gives constant acceleration, while \(\vec{B}\) bends the trajectory. Motion is confined to the plane formed by directions of \(\vec{E}\) and \(\vec{v}\times \vec{B}\).
Updated On: Aug 28, 2025
  • The charge will move in the direction of \(z\) with constant velocity.
  • The charge will always move on the \(y\text{-}z\) plane only.
  • The trajectory of the charge will be a circle.
  • The charge will progress in the direction of \(y\).
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The Correct Option is B, D

Solution and Explanation

Step 1: Initial acceleration.
At \(t=0\), the particle is at rest. Electric force acts: \[ \vec{F}_E = q \vec{E} \] Since \(\vec{E}\) is along \(z\), the charge accelerates in the \(+z\) direction.

Step 2: Effect of magnetic field.
As soon as the charge gains velocity \(\vec{v}_z\), the magnetic force acts: \[ \vec{F}_B = q (\vec{v} \times \vec{B}) \] With \(\vec{v}\) along \(z\) and \(\vec{B}\) along \(x\): \[ \vec{v}_z \times \vec{B}_x \; \Rightarrow \; \vec{F}_B \; \text{along } y \] Thus, the charge is deflected into the \(y\)-direction.

Step 3: Motion confined to plane.
Velocity components exist only in \(z\) and \(y\), hence motion is restricted to the \(y\)-\(z\) plane.

Step 4: Nature of trajectory.
The trajectory is not circular, because there is a continuous acceleration along \(z\) due to constant \(E\). Thus, the particle drifts along \(y\) while being accelerated in \(z\). Conclusions: - (A) is false (velocity along \(z\) is not constant, it increases). - (B) is true (motion confined to \(y\)-\(z\) plane). - (C) is false (not circular). - (D) is true (charge progresses in \(y\)-direction due to Lorentz force).

Final Answer:
\[ \boxed{(B) \; \text{and} \; (D)} \]

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