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

The correct answer is option (A): Circular path, \(\frac{1}{2}\)sec.

Answer 2

To solve this problem, we need to analyze the motion of a charged particle in a uniform magnetic field. The particle starts from the origin with an initial velocity and follows a specific trajectory due to the Lorentz force.
Given:
- Magnetic field \(\vec{B} = B_0 \hat{k}\)
- Initial velocity \(\vec{v} = 3 \hat{i} + 4 \hat{k}\) m/s
Solution:
1. Trajectory of the Particle
The motion of a charged particle in a magnetic field is determined by the Lorentz force:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
Since the magnetic field is in the \(\hat{k}\) direction (\(\vec{B} = B_0 \hat{k}\)), we can find the force acting on the particle due to this field. The initial velocity has components in the \(\hat{i}\) and \(\hat{k}\) directions:
\[ \vec{v} = 3 \hat{i} + 4 \hat{k} \]
The cross product \(\vec{v} \times \vec{B}\) gives us the direction of the force:
\[ \vec{v} \times \vec{B} = (3 \hat{i} + 4 \hat{k}) \times (B_0 \hat{k}) \]
Using the properties of the cross product:
\[ \hat{i} \times \hat{k} = -\hat{j} \]
\[ \hat{k} \times \hat{k} = 0 \]
So,
\[ \vec{v} \times \vec{B} = 3 \hat{i} \times B_0 \hat{k} + 4 \hat{k} \times B_0 \hat{k} = 3 B_0 (-\hat{j}) + 0 = -3 B_0 \hat{j} \]
The force is perpendicular to both \(\vec{v}\) and \(\vec{B}\), indicating circular motion in the plane perpendicular to \(\vec{B}\).
Therefore, the trajectory of the particle is a circular path in the \(xy\)-plane.
 2. Time to Reach 2m Above the \(xy\)-Plane
The initial velocity component in the \(\hat{k}\) direction is \(4\) m/s, and there is no force acting in the \(z\)-direction since the magnetic field only affects the perpendicular velocity components. Hence, the motion in the \(z\)-direction is uniform.
The displacement in the \(z\)-direction can be given by:
\[ z = v_z t \]
Given \(z = 2\) m and \(v_z = 4\) m/s:
\[ 2 = 4t \]
\[ t = \frac{2}{4} = \frac{1}{2} \text{ seconds} \]
Thus, the particle reaches 2 meters above the \(xy\)-plane in \(\frac{1}{2}\) seconds.
Conclusion
- The trajectory of the particle is a circular path.
- The time \(t\) at which it reaches 2 meters above the \(xy\)-plane is \(\frac{1}{2}\) seconds.
Therefore, the correct answer is:
- (A): Circular path, \(\frac{1}{2}\) seconds.