To determine the position vector of a charged particle moving in a uniform magnetic field, we need to analyze the motion of the particle under the influence of the magnetic force. The particle's motion can be described using the Lorentz force law, which states that the force acting on a charged particle in a magnetic field is given by the equation: F = q(\mathbf{u} \times \mathbf{B}). Here, q is the charge of the particle, u is its velocity, and B is the magnetic field.
Understanding the Motion
In this scenario, the magnetic field is uniform and directed along the z-axis, represented as B = B_0 \hat{k}. The velocity of the particle at time t = 0 is given by \mathbf{u} = u_x \hat{i} + u_z \hat{k}. The presence of the magnetic field will cause the particle to experience a centripetal force, resulting in circular motion in the plane perpendicular to the magnetic field.
Force and Acceleration
The magnetic force acting on the particle can be calculated as follows:
- F = q(\mathbf{u} \times \mathbf{B})
- F = q[(u_x \hat{i} + u_z \hat{k}) \times (B_0 \hat{k})]
- F = qB_0 u_x \hat{j}
This force acts in the y-direction, causing the particle to move in a circular path in the x-y plane while maintaining its z-component of velocity constant.
Determining the Circular Motion
The radius of the circular motion can be found using the relationship between the centripetal force and the magnetic force:
F = m \frac{v^2}{r} = qB_0 v
From this, we can derive the radius r of the circular path:
r = \frac{mv}{qB_0}
In our case, the speed in the x-direction is u_x, so:
r = \frac{mu_x}{qB_0}
Time Period of Motion
The time period T for one complete revolution can be calculated as:
T = \frac{2\pi r}{v} = \frac{2\pi m}{qB_0}
Since the particle is moving in a circular path, we can find the angular displacement after a time t = \frac{9\pi m}{qB_0}:
\theta = \frac{t}{T} \cdot 2\pi = \frac{9\pi m}{qB_0} \cdot \frac{qB_0}{2\pi m} = \frac{9}{2} \cdot 2\pi = 9\pi
Position Vector Calculation
After a time of 9\pi m / qB_0, the particle completes 4.5 full revolutions (since 9\pi / 2\pi = 4.5). This means it returns to the same position in the x-y plane but will have moved in the z-direction due to its constant velocity u_z.
The displacement in the z-direction after this time is:
z = u_z \cdot t = u_z \cdot \frac{9\pi m}{qB_0}
In the x-y plane, after 4.5 revolutions, the particle will be at the negative y-axis (since it starts at the origin and moves counterclockwise). The displacement in the y-direction is:
y = -2r = -2 \cdot \frac{mu_x}{qB_0}
Final Position Vector
Combining these results, the position vector \mathbf{r} at time t = \frac{9\pi m}{qB_0} is:
\mathbf{r} = -2 \frac{mu_x}{qB_0} \hat{j} + 9\frac{u_z \pi m}{qB_0} \hat{k}
Thus, the correct answer is:
B. \mathbf{r} = -2 \frac{mu_x}{qB_0} \hat{j} + 9\frac{u_z \pi m}{qB_0} \hat{k}
