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The problem requires us to calculate the magnetic force acting on an electron and a proton moving through a magnetic field. We'll use the formula for the magnetic force on a charged particle:

\[
F = |q| \cdot |\vec{v} \times \vec{B}|
\]

Where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle (for an electron, \( q_e = -1.6 \times 10^{-19} \, \text{C} \); for a proton, \( q_p = +1.6 \times 10^{-19} \, \text{C} \)),
- \( \vec{v} \) is the velocity vector of the particle,
- \( \vec{B} \) is the magnetic field vector,
- \( \times \) denotes the cross product.

### Given:
- Velocity vector of the electron: \( \vec{v} = (2.7 \times 10^6 \, \text{m/s}, 4.3 \times 10^6 \, \text{m/s}) \),
- Magnetic field vector: \( \vec{B} = (0.023 \, \text{T}, -0.12 \, \text{T}) \).

Now, we'll compute the cross product \( \vec{v} \times \vec{B} \):

The cross product in two dimensions (with \( z = 0 \)) for vectors \( \vec{v} = (v_x, v_y) \) and \( \vec{B} = (B_x, B_y) \) is:

\[
\vec{v} \times \vec{B} = v_x \cdot B_y - v_y \cdot B_x
\]

Substituting the given values:

\[
\vec{v} \times \vec{B} = (2.7 \times 10^6) \cdot (-0.12) - (4.3 \times 10^6) \cdot (0.023)
\]

\[
\vec{v} \times \vec{B} = -3.24 \times 10^5 - 9.89 \times 10^4
\]

\[
\vec{v} \times \vec{B} = -4.229 \times 10^5 \, \text{m}^2/\text{s}^2
\]

### (a) For the electron:
Now, calculate the force using the magnitude of the electron's charge:

\[
F_e = |q_e| \cdot |\vec{v} \times \vec{B}| = 1.6 \times 10^{-19} \cdot 4.229 \times 10^5
\]

\[
F_e = 6.766 \times 10^{-14} \, \text{N}
\]

### (b) For the proton:
For the proton, the same calculation applies because the magnitude of the charge is the same (just positive), so:

\[
F_p = 6.766 \times 10^{-14} \, \text{N}
\]

Thus, the magnitude of the magnetic force on both the electron and the proton is approximately \( 6.77 \times 10^{-14} \, \text{N} \).

Would you like any further details on these steps or more clarification? 

### Related Questions:
1. How is the direction of the magnetic force determined for a proton compared to an electron?
2. What happens to the magnetic force if the velocity of the particle is perpendicular to the magnetic field?
3. How would the force change if the magnetic field strength is doubled?
4. How do you determine the direction of the force using the right-hand rule?
5. What role does the sign of the charge play in the direction of the force?

### Tip:
Remember, the direction of the magnetic force is determined by the cross product, so for negatively charged particles (like the electron), the force will be in the opposite direction compared to positively charged particles.

An electron that has a velocity with x component 2.7 × 10^6 m/s and y component 4.3 × 10^6 m/s moves through a uniform magnetic field with x component 0.023 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Calculate the magnetic force on an electron and proton with specific velocity components moving through a uniform magnetic field using the Lorentz force law. This problem involves vector calculus and electromagnetism principles, suitable for college-level physics students.

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