To solve the problem, we need to analyze the motion of the charged particle in the magnetic field and the conditions for the collision with the uncharged particle.
### Step 1: Determine the force acting on the charged particle
The charged particle experiences a magnetic force given by the Lorentz force equation:
\[
\vec{F} = q(\vec{v} \times \vec{B})
\]
where \( q = 10 \mu C = 10 \times 10^{-6} C \), \( \vec{v} = (10 \, \text{cm}) \, \hat{i} + (20 \, \text{cm}) \, \hat{j} = (0.1 \, \text{m}) \, \hat{i} + (0.2 \, \text{m}) \, \hat{j} \), and \( \vec{B} = 0.6 \pi \, T \, \hat{i} \).
### Step 2: Calculate the magnetic force
First, we need to compute the cross product \( \vec{v} \times \vec{B} \):
\[
\vec{v} \times \vec{B} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0.1 & 0.2 & 0 \\
0.6\pi & 0 & 0
\end{vmatrix}
\]
Calculating the determinant, we find:
\[
\vec{v} \times \vec{B} = (0.2 \cdot 0 - 0 \cdot 0) \hat{i} - (0.1 \cdot 0 - 0 \cdot 0.6\pi) \hat{j} + (0.1 \cdot 0 - 0.2 \cdot 0.6\pi) \hat{k} = -0.12\pi \hat{k}
\]
Thus,
\[
\vec{F} = q(-0.12\pi \hat{k}) = -1.2 \times 10^{-6} \pi \hat{k} \, N
\]
### Step 3: Determine the acceleration of the charged particle
Using Newton's second law, \( \vec{F} = m \vec{a} \):
\[
\vec{a} = \frac{\vec{F}}{m} = \frac{-1.2 \times 10^{-6} \pi \hat{k}}{3 \times 10^{-6}} = -0.4\pi \hat{k} \, m/s^2
\]
### Step 4: Find the radius of the circular path
The charged particle will move in a circular path due to the magnetic force. The radius \( r \) of the circular motion can be found using:
\[
r = \frac{mv}{qB}
\]
Calculating \( v \):
\[
v = |\vec{v}| = \sqrt{(0.1)^2 + (0.2)^2} = \sqrt{0.01 + 0.04} = \sqrt{0.05} = 0.2236 \, m/s
\]
Now substituting the values:
\[
r = \frac{(3 \times 10^{-6}) (0.2236)}{(10 \times 10^{-6})(0.6\pi)} = \frac{0.0006708}{0.0001884\pi} \approx \frac{0.0006708}{0.000592} \approx 1.13 \, m
\]
### Step 5: Determine the conditions for collision
The uncharged particle is moving along the negative x-axis with a constant velocity. For the two particles to collide, they must meet at the same point at the same time. The charged particle starts at the origin and moves in a circular path with radius \( r \).
### Step 6: Set up the equations for collision
Let the mass of the uncharged particle be \( M \) and its velocity be \( v_u \) (along negative x-axis). The position of the uncharged particle at time \( t \) is:
\[
x_u = 50 \, cm - v_u t
\]
The charged particle moves in a circular path, which can be described parametrically. The position of the charged particle at time \( t \) can be described as:
\[
x_c = r \cos(\omega t), \quad y_c = r \sin(\omega t)
\]
where \( \omega = \frac{qB}{m} \).
### Step 7: Solve for \( M \)
The two particles collide when:
\[
r \cos(\omega t) = 50 - v_u t
\]
\[
r \sin(\omega t) = 0
\]
This implies \( \sin(\omega t) = 0 \), leading to \( \omega t = n\pi \) for \( n = 0, 1, 2, \ldots \).
### Step 8: Calculate the mass of the uncharged particle
Using conservation of momentum before and after the collision:
\[
mv + Mu = (m + M)v_f
\]
where \( v_f \) is the final velocity after collision. The final velocity will be the same as the velocity of the charged particle at the collision point.
### Conclusion
The mass \( M \) of the uncharged particle can be calculated based on the above equations and conditions for collision.
