A particle of mass 6.0 g moves at 4.0 km`//`s in an xy plane, in a region with a uniform magnetic field given by 5.0 mT. At one instatnt, when the particles velocity is directed `37^(@)` counterclockwise from the positive direction of the x-axis, the magnetic force on the particle is `0.45hatk` N. What is the particle's charge?

Answer

To find the charge of the particle, we can use the formula for the magnetic force acting on a charged particle moving in a magnetic field. The magnetic force \( \mathbf{F} \) is given by:

\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]

Where:
- \( \mathbf{F} \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( \mathbf{v} \) is the velocity vector of the particle,
- \( \mathbf{B} \) is the magnetic field vector.

### Step 1: Convert the given quantities into consistent units
- Mass of the particle: \( m = 6.0 \, \text{g} = 6.0 \times 10^{-3} \, \text{kg} \)
- Velocity: \( v = 4.0 \, \text{km/s} = 4.0 \times 10^3 \, \text{m/s} \)
- Magnetic field: \( B = 5.0 \, \text{mT} = 5.0 \times 10^{-3} \, \text{T} \)

### Step 2: Write the velocity vector in component form
The velocity is directed at an angle of \( 37^\circ \) counterclockwise from the positive x-axis. Therefore, we can express the velocity vector \( \mathbf{v} \) as:

\[ \mathbf{v} = v \cos(37^\circ) \hat{i} + v \sin(37^\circ) \hat{j} \]

Calculating the components:
- \( \cos(37^\circ) \approx 0.8 \)
- \( \sin(37^\circ) \approx 0.6 \)

Thus,

\[ \mathbf{v} = (4.0 \times 10^3)(0.8) \hat{i} + (4.0 \times 10^3)(0.6) \hat{j} = 3200 \hat{i} + 2400 \hat{j} \, \text{m/s} \]

### Step 3: Write the magnetic field vector
The magnetic field is along the positive x-axis:

\[ \mathbf{B} = 5.0 \times 10^{-3} \hat{i} \, \text{T} \]

### Step 4: Calculate the cross product \( \mathbf{v} \times \mathbf{B} \)
Using the determinant method for the cross product:

\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3200 & 2400 & 0 \\ 5.0 \times 10^{-3} & 0 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \left( 2400 \cdot 0 - 0 \cdot 0 \right) - \hat{j} \left( 3200 \cdot 0 - 0 \cdot 5.0 \times 10^{-3} \right) + \hat{k} \left( 3200 \cdot 0 - 2400 \cdot 5.0 \times 10^{-3} \right) \]

This simplifies to:

\[ = 0 \hat{i} - 0 \hat{j} - 2400 \cdot 5.0 \times 10^{-3} \hat{k} = -12 \hat{k} \, \text{m}^2/\text{s} \]

### Step 5: Substitute into the magnetic force equation
Given that the magnetic force \( \mathbf{F} = 0.45 \hat{k} \, \text{N} \):

\[ 0.45 \hat{k} = q (-12 \hat{k}) \]

### Step 6: Solve for the charge \( q \)
From the equation above, we can isolate \( q \):

\[ q = \frac{0.45}{-12} \]

Calculating \( q \):

\[ q = -0.0375 \, \text{C} = -3.75 \times 10^{-2} \, \text{C} \]

### Final Answer
The charge of the particle is:

\[ q = -3.75 \times 10^{-2} \, \text{C} \]