To solve the problem, we will follow these steps:
### Step 1: Understand the initial conditions
The particle has a mass \( m \), charge \( q \), and is projected with a velocity \( v \) at an angle \( \theta \) with respect to a uniform magnetic field \( B \). The magnetic field is assumed to be along the \( z \)-axis (or \( k \)-direction).
### Step 2: Resolve the initial velocity
The initial velocity \( \vec{v} \) can be resolved into two components:
- The component parallel to the magnetic field (along the \( z \)-axis):
\[
v_z = v \cos \theta
\]
- The component perpendicular to the magnetic field (in the \( y \)-direction):
\[
v_y = v \sin \theta
\]
Thus, the initial velocity vector can be expressed as:
\[
\vec{v}_0 = v \sin \theta \, \hat{j} + v \cos \theta \, \hat{k}
\]
### Step 3: Analyze the motion in the magnetic field
In a magnetic field, the force on a charged particle is given by the Lorentz force:
\[
\vec{F} = q (\vec{v} \times \vec{B})
\]
Since the magnetic field is along the \( z \)-axis, the force will act in the \( x \)- and \( y \)-directions, causing circular motion in the plane perpendicular to the magnetic field.
### Step 4: Determine the time of motion
The time given in the problem is:
\[
t = \frac{\pi m}{2qB}
\]
This time corresponds to a quarter of the period of revolution of the particle in the magnetic field.
### Step 5: Find the final velocity after time \( t \)
After a time \( t \), the particle will have rotated \( 90^\circ \) in the plane perpendicular to the magnetic field. The \( z \)-component of the velocity remains unchanged:
\[
v_z = v \cos \theta
\]
The \( y \)-component of the velocity will change direction:
\[
v_y' = -v \sin \theta
\]
Thus, the final velocity vector after time \( t \) is:
\[
\vec{v}_f = -v \sin \theta \, \hat{i} + v \cos \theta \, \hat{k}
\]
### Step 6: Calculate the change in velocity
The change in velocity \( \Delta \vec{v} \) is given by:
\[
\Delta \vec{v} = \vec{v}_f - \vec{v}_0
\]
Substituting the values:
\[
\Delta \vec{v} = \left(-v \sin \theta \, \hat{i} + v \cos \theta \, \hat{k}\right) - \left(v \sin \theta \, \hat{j} + v \cos \theta \, \hat{k}\right)
\]
This simplifies to:
\[
\Delta \vec{v} = -v \sin \theta \, \hat{i} - v \sin \theta \, \hat{j}
\]
### Step 7: Calculate the magnitude of the change in velocity
The magnitude of the change in velocity is:
\[
|\Delta \vec{v}| = \sqrt{(-v \sin \theta)^2 + (-v \sin \theta)^2} = \sqrt{2(v \sin \theta)^2} = v \sin \theta \sqrt{2}
\]
### Final Answer
The magnitude of the change in velocity of the particle after time \( t \) is:
\[
|\Delta \vec{v}| = v \sqrt{2} \sin \theta
\]
