Question

A positively charged particle of mass m is passed through a velocity selector. It moves horizontally rightward without deviation along the line y = \(\frac{2mv}{qB}\) with a speed v. The electric field is vertically downwards and magnetic field is into the plane of the paper. Now, the electric field is switched off at t = 0. The angular momentum of the charged particle about origin O at t = \(\frac{\pi m}{qB}\) is

• A. \(\frac{2mE^2}{qB^3}\)
• B. zero
• C. \(\frac{mE^3}{qB^2}\)
• D. \(\frac{mE^2}{qB^3}\)

Answer 1

The Correct Option is B

Step 1: Understanding the scenario 
Initially, the particle moves horizontally with constant velocity \( v \) due to the velocity selector. The electric field (\( E \)) is vertically downward, and the magnetic field (\( B \)) is into the plane of the paper. At time \( t=0 \), the electric field is switched off, leaving only the magnetic field acting on the particle.

Step 2: Radius of the circular path after switching off the electric field
After the electric field is switched off, the particle moves under the influence of only the magnetic field, thus describing a circular path. The radius (\( r \)) of this path is given by:

\[ r = \frac{mv}{qB} \]

Step 3: Identify particle position at \( t = \frac{\pi m}{qB} \)
The time period \( T \) of circular motion under a magnetic field is given by:

\[ T = \frac{2\pi m}{qB} \]

Thus, the given time \( t = \frac{\pi m}{qB} \) is exactly half of the time period (\( T/2 \)). At half the time period, the particle will have rotated exactly 180° from its original direction.

This means at \( t = \frac{\pi m}{qB} \), the particle will be horizontally opposite to the original position, now traveling towards the left.

Step 4: Position of the particle at the given time
Initially, particle was moving horizontally rightward along the line:

\[ y = \frac{2mv}{qB} \]

This indicates the particle was initially located at a vertical distance of \( 2r \) from the origin (since \( r = \frac{mv}{qB} \), thus \( y = 2r \)).
Thus, the particle started at the topmost point of a circle of radius \( r \).

After half rotation (180°), the particle will be exactly at the bottom point of the circle, located at the same vertical distance below the initial line (thus, on the horizontal axis, \( y=0 \)). The particle is now located exactly on the x-axis, passing through the origin.

At this moment, the particle’s velocity vector is horizontal to the left.

Step 5: Calculating angular momentum about origin O at this moment
Angular momentum (\( L \)) about any point is given by:

\[ \vec{L} = \vec{r} \times (m\vec{v}) \]

Here, the position vector \( \vec{r} \) passes through the origin (since particle is exactly on the x-axis at the bottom of the circle), making the particle's position vector and velocity vector exactly collinear (position vector along x-axis and velocity along negative x-axis).

The cross product of two collinear vectors is always zero:

\[ \vec{L} = \vec{r} \times m\vec{v} = 0 \]

Thus, the angular momentum about the origin O at this moment is zero.

Answer 2

At \( t = 0 \), the electric field is turned off, and only the magnetic field is present. The velocity selector ensures that the charged particle moves along the \(y = \dfrac{2mv}{qB}\) line without deviation. This means the force due to the electric field and magnetic field must cancel each other out, so: \[ \vec{E} = \dfrac{vB}{q} \hat{j} \] Since the electric field has been turned off, the particle is only under the influence of the magnetic field. The magnetic force is acting perpendicular to the velocity vector, meaning the particle will move in a circular path. As the velocity vector is tangent to the circular path, there is no component of the velocity that leads to a change in the position along the center of the motion.

Thus, the angular momentum of the charged particle about the origin \(O\) is zero at the time \(t = 0\).

Thus, the correct answer is (B) zero.