Question

Arrange the following steps in the chronological order, when a charged particle enters perpendicularly into a uniform magnetic field. (A) The charged particle starts moving in a circular path. (B) Net work done by the field is zero. (C) The speed of the charged particle remains constant but its direction changes. (D) A force acts perpendicular to both the velocity of the charged particle and the magnetic field. Choose the correct answer from the options given below:

• A. (D), (A), (C), (B)
• B. (A), (C), (D), (B)
• C. (C), (A), (D), (B)
• D. (A), (C), (B), (D)

Hint:

For motion of a charged particle in a uniform magnetic field: \[ \vec F=q(\vec v\times\vec B) \] \[ F \perp v \] \[ \text{Speed remains constant} \] \[ \text{Direction changes continuously} \] \[ \text{Work done by magnetic field}=0 \] Hence the particle moves in a circular path.

Answer 1

The Correct Option is A

Step 1:Identify the first event. As soon as the particle enters the magnetic field, \[ {\text{A magnetic force acts on the particle.}} \] The force is perpendicular to both velocity and magnetic field. Hence, \[ (D) \] occurs first.

Step 2: Determine the next consequence. Because the force is always perpendicular to velocity, it acts as a centripetal force. Therefore, \[ {\text{The particle starts moving in a circular path.}} \] Hence, \[ (A) \] occurs next.

Step 3: Analyse the motion. Since the force is perpendicular to velocity, \[ \vec F\cdot\vec v=0 \] Thus only the direction changes while the speed remains constant. \[ {\text{Speed remains constant but direction changes.}} \] Hence, \[ (C) \] follows.

Step 4: Determine the work done. Since the magnetic force is always perpendicular to displacement, \[ W=\int \vec F\cdot d\vec r=0 \] Therefore, \[ {\text{Net work done by the magnetic field is zero.}} \] Hence, \[ (B) \] is the final statement.

Step 5: Write the chronological order. \[ (D)\rightarrow(A)\rightarrow(C)\rightarrow(B) \] \[ { (D),\ (A),\ (C),\ (B) } \] Hence, the correct option is \[ {(A)} \]