Let's re-examine the calculation of the $x$-component of the velocity ($v_x$) while carefully considering the effect of the negative charge.

Given Data:

• Charge of the particle: $q = -5.20 \times 10^{-9} \, \text{C}$
• Magnetic field: $\vec{B} = -1.25 \, \text{T} \, \hat{k}$
• Magnetic force: $\vec{F} = -4.00 \times 10^{-7} \, \text{N} \, \hat{i} + 7.60 \times 10^{-7} \, \text{N} \, \hat{j}$

Magnetic Force Equation:

The force on the particle due to the magnetic field is given by:

$\vec{F} = q \, \vec{v} \times \vec{B}$

Force in the $y$-Direction:

The $y$-component of the magnetic force is related to the $x$-component of the velocity:

$F_y = -q \, v_x \, B_z$

Solving for $v_x$:

$v_x = -\frac{F_y}{q \, B_z}$

Substituting the values:

$v_x = -\frac{7.60 \times 10^{-7} \, \text{N}}{(-5.20 \times 10^{-9} \, \text{C}) \times (-1.25 \, \text{T})}$

Simplifying further:

$v_x = -\frac{7.60 \times 10^{-7}}{6.5 \times 10^{-9}} \, \text{m/s}$

Calculate $v_x$:

$v_x \approx -117 \, \text{m/s}$

So, the correct $x$-component of the velocity is $\boxed{-117 \, \text{m/s}}$.

This negative sign indicates that the velocity in the $x$-direction is opposite to what we initially assumed when considering the force's direction.

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Would you like to explore more details or have additional questions?

Related Questions:

1. What is the general direction of the velocity when the magnetic force is given?
2. How would the velocity components change if the magnetic field direction were reversed?
3. What is the significance of the sign of the charge on the particle?
4. Can you explain the cross-product rule used in calculating magnetic force?
5. How does the magnetic force vary with different angles between velocity and magnetic field?

Tip: When dealing with vector cross-products, always be mindful of the charge's sign, as it influences the direction of the resulting force.