To solve the problem of finding the Lorentz force on a proton moving in an electric and magnetic field, we will follow these steps:
### Step 1: Identify the Given Values
- **Velocity of the proton (v)**: \(5 \times 10^5 \, \text{m/s}\) (along the Y-direction)
- **Electric field (E)**: \(10^5 \, \text{V/m}\) (along the X-direction)
- **Magnetic field (B)**: \(1 \, \text{Wb/m}^2\) (along the Z-direction)
- **Charge of the proton (Q)**: \(1.6 \times 10^{-19} \, \text{C}\)
### Step 2: Calculate the Electric Force (F_E)
The electric force on the proton can be calculated using the formula:
\[
F_E = Q \cdot E
\]
Substituting the values:
\[
F_E = (1.6 \times 10^{-19} \, \text{C}) \cdot (10^5 \, \text{V/m}) = 1.6 \times 10^{-14} \, \text{N}
\]
The direction of this force is along the X-direction (positive I direction).
### Step 3: Calculate the Magnetic Force (F_B)
The magnetic force can be calculated using the formula:
\[
F_B = Q \cdot (v \times B)
\]
First, we need to determine the cross product \(v \times B\). The velocity vector \(v\) is in the Y-direction, and the magnetic field vector \(B\) is in the Z-direction:
- \(v = 5 \times 10^5 \, \hat{j}\)
- \(B = 1 \, \hat{k}\)
Using the right-hand rule for the cross product:
\[
v \times B = \hat{j} \times \hat{k} = \hat{i}
\]
Thus,
\[
F_B = Q \cdot (v \times B) = Q \cdot (5 \times 10^5 \, \hat{i})
\]
Substituting the values:
\[
F_B = (1.6 \times 10^{-19} \, \text{C}) \cdot (5 \times 10^5 \, \hat{i}) = 8.0 \times 10^{-14} \, \hat{i} \, \text{N}
\]
### Step 4: Calculate the Total Lorentz Force (F)
The total Lorentz force is the sum of the electric force and the magnetic force:
\[
F = F_E + F_B
\]
Substituting the forces:
\[
F = (1.6 \times 10^{-14} \, \hat{i}) + (8.0 \times 10^{-14} \, \hat{i}) = 9.6 \times 10^{-14} \, \hat{i} \, \text{N}
\]
### Final Answer
The Lorentz force on the proton is:
\[
F = 9.6 \times 10^{-14} \, \text{N} \, \hat{i}
\]
This indicates that the force is directed along the positive X-direction.
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