Question

A charged particle in a uniform magnetic field \(\vec{B}=B_0\hat{k}\) starts moving from the origin with velocity \(v=3\hat{i}+4\hat{k}\) m/s.The trajectory of the particle and the time \(t\) at which it reaches 2m above the x-y plane are 

• A. Helical path, \(\frac{1}{2}\) sec.
• B. Circular path, \(\frac{1}{2}\)sec.
• C. Circular path, \(\frac{2}{3}\)sec.
• D. Helical path, \(\frac{2}{3}\) sec.

Answer 1

The Correct Option is A

Given: 
Magnetic field: $\vec{B} = B_0 \hat{k}$
Initial velocity: $\vec{v} = 3 \hat{i} + 4 \hat{k}$ m/s

Explanation:
The velocity has two components:
- Perpendicular to $\vec{B}$: $3 \hat{i}$ → causes circular motion in the x-y plane
- Parallel to $\vec{B}$: $4 \hat{k}$ → causes linear motion along the z-axis

Therefore, the trajectory is a helical path.

To find the time when the particle is 2 m above the x-y plane:
Vertical velocity: $v_z = 4$ m/s
Displacement in z-direction: $z = 2$ m
Using $z = v_z \cdot t$:
$2 = 4t \Rightarrow t = \frac{1}{2}$ sec

Final Answer:
Helical path, $t = \frac{1}{2}$ sec
Correct option: (B)

Answer 2

Given:
- Magnetic field: $\vec{B} = B_0 \hat{k}$
- Initial velocity: $\vec{v} = 3 \hat{i} + 4 \hat{k}$ m/s 

1. Trajectory of the Particle:
The force on the charged particle is given by:
$\vec{F} = q (\vec{v} \times \vec{B})$
$\vec{v} = 3\hat{i} + 4\hat{k}$, and $\vec{B} = B_0 \hat{k}$
So,
$\vec{v} \times \vec{B} = (3\hat{i} + 4\hat{k}) \times B_0\hat{k} = 3B_0(-\hat{j}) + 0 = -3B_0 \hat{j}$
This force is perpendicular to $\vec{B}$ and causes circular motion in the $xy$-plane.
But since there is also a constant velocity component along $\hat{k}$, the particle follows a helical path.

2. Time to Reach 2 m Above the $xy$-Plane:
Motion along the $z$-axis is uniform with $v_z = 4$ m/s:
$z = v_z \cdot t$
$2 = 4t \Rightarrow t = \frac{1}{2}$ sec

Final Answer:
Helical path, $t = \frac{1}{2}$ seconds
Correct option: (B)