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:
\[
\boxed{(B) \; \text{and} \; (D)}
\]
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in V/Ω, is __________ (round off to two decimal places).
The relationship between two variables \( x \) and \( y \) is given by \( x + py + q = 0 \) and is shown in the figure. Find the values of \( p \) and \( q \). Note: The figure shown is representative.