A proton moves at a constant velocity of `50 m/s` along the x-axis, in uniform electric and magnetic fields. The magnetic field is `B = (2.0 mT)hatj`. What is the electric field?

Explanation

To find the electric field when a proton moves at a constant velocity in the presence of uniform electric and magnetic fields, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Velocity of the proton, \( v = 50 \, \text{m/s} \) (along the x-axis).
- Magnetic field, \( B = 2.0 \, \text{mT} = 2.0 \times 10^{-3} \, \text{T} \) (along the y-axis).

2. **Understand the Motion:**
- The proton is moving with constant velocity, which implies that the net force acting on it is zero. This means that the magnetic force and the electric force must balance each other.

3. **Calculate the Magnetic Force:**
- The magnetic force \( F_B \) on a charged particle moving in a magnetic field is given by the equation:
\[ F_B = q(v \times B) \]
- For a proton, the charge \( q \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
- The velocity vector \( \vec{v} \) can be represented as \( 50 \hat{i} \, \text{m/s} \) and the magnetic field vector \( \vec{B} \) as \( 2.0 \times 10^{-3} \hat{j} \, \text{T} \).
- The cross product \( v \times B \) can be calculated using the determinant:
\[ v \times B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 50 & 0 & 0 \\ 0 & 2.0 \times 10^{-3} & 0 \end{vmatrix} \]
- This results in:
\[ v \times B = (50 \cdot 2.0 \times 10^{-3}) \hat{k} = 0.1 \hat{k} \, \text{m}^2/\text{s}^2 \]
- Therefore, the magnetic force is:
\[ F_B = q \cdot (0.1 \hat{k}) = (1.6 \times 10^{-19}) \cdot (0.1) \hat{k} = 1.6 \times 10^{-20} \hat{k} \, \text{N} \]

4. **Set Up the Force Balance:**
- Since the net force is zero, the electric force \( F_E \) must equal the magnetic force in magnitude but act in the opposite direction:
\[ F_E = -F_B \]
- Thus, the electric force is:
\[ F_E = -1.6 \times 10^{-20} \hat{k} \, \text{N} \]

5. **Calculate the Electric Field:**
- The electric force is also given by:
\[ F_E = qE \]
- Setting the two expressions for electric force equal gives:
\[ qE = -1.6 \times 10^{-20} \hat{k} \]
- Solving for \( E \):
\[ E = \frac{-1.6 \times 10^{-20}}{1.6 \times 10^{-19}} = -0.1 \hat{k} \, \text{N/C} \]
- Therefore, the electric field is:
\[ E = -0.1 \hat{k} \, \text{N/C} \quad \text{or} \quad E = -0.1 \hat{k} \, \text{V/m} \]

### Final Answer:
The electric field is \( E = -0.1 \hat{k} \, \text{V/m} \).