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Sagot :
Sure! Let's take this step-by-step to solve the problem for each part:
### Part (a): Calculate the total force on the moving particle
Given:
- Charge of the particle, [tex]\( q = 3.68 \times 10^{-18} \, \text{C} \)[/tex]
- Velocity of the particle, [tex]\( \vec{v} = 4 \hat{i} + 4 \hat{j} - \hat{k} \, \text{m/s} \)[/tex]
- Magnetic field, [tex]\( \vec{B} = 4 \hat{i} + 3 \hat{j} + \hat{k} \, \text{T} \)[/tex]
- Electric field, [tex]\( \vec{E} = 3 \hat{i} - \hat{j} - 2 \hat{k} \, \text{V/m} \)[/tex]
To determine the total force on the charged particle, we need to calculate both the electric force ([tex]\(\vec{F}_e\)[/tex]) and the magnetic force ([tex]\(\vec{F}_b\)[/tex]).
The electric force is given by:
[tex]\[ \vec{F}_e = q \vec{E} \][/tex]
The magnetic force is given by:
[tex]\[ \vec{F}_b = q (\vec{v} \times \vec{B}) \][/tex]
The total force is:
[tex]\[ \vec{F} = \vec{F}_e + \vec{F}_b \][/tex]
Given results for the force components:
[tex]\[ F_x = 3.68 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_y = -3.3119999999999996 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_z = -2.208 \times 10^{-17} \, \text{N} \][/tex]
### Part (b): Angle with the [tex]\( +x \)[/tex]-axis
To find the angle the force vector makes with the positive [tex]\( x \)[/tex]-axis, we use the dot product formula between the force vector and the [tex]\( x \)[/tex]-axis unit vector:
[tex]\[ \theta = \cos^{-1} \left( \frac{\vec{F} \cdot \hat{i}}{|\vec{F}|} \right) \][/tex]
Where
[tex]\[ \vec{F} \cdot \hat{i} = F_x \][/tex]
[tex]\[ |\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} \][/tex]
Given result for the angle:
[tex]\[ \theta \approx 47.24660675818882^\circ \][/tex]
### Part (c): Electric field for zero net force
For the total force on the particle to be zero, the electric force must exactly cancel out the magnetic force:
[tex]\[ \vec{F}_e = -\vec{F}_b \][/tex]
[tex]\[ q \vec{E}_{\text{req}} = -q (\vec{v} \times \vec{B}) \][/tex]
[tex]\[ \vec{E}_{\text{req}} = - (\vec{v} \times \vec{B}) / q \][/tex]
Given results for the required electric field components:
[tex]\[ E_x = -7.0 \, \text{V/m} \][/tex]
[tex]\[ E_y = 8.0 \, \text{V/m} \][/tex]
[tex]\[ E_z = 4.0 \, \text{V/m} \][/tex]
### Summary of Answers
(a) The components of the total force are:
[tex]\[ F_x = 3.68 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_y = -3.3119999999999996 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_z = -2.208 \times 10^{-17} \, \text{N} \][/tex]
(b) The angle the force vector makes with the positive [tex]\( x \)[/tex]-axis is approximately [tex]\( 47.25^\circ \)[/tex] counterclockwise from the [tex]\( +x \)[/tex]-axis.
(c) The components of the electric field that would make the total force on the particle zero are:
[tex]\[ E_x = -7.0 \, \text{V/m} \][/tex]
[tex]\[ E_y = 8.0 \, \text{V/m} \][/tex]
[tex]\[ E_z = 4.0 \, \text{V/m} \][/tex]
### Part (a): Calculate the total force on the moving particle
Given:
- Charge of the particle, [tex]\( q = 3.68 \times 10^{-18} \, \text{C} \)[/tex]
- Velocity of the particle, [tex]\( \vec{v} = 4 \hat{i} + 4 \hat{j} - \hat{k} \, \text{m/s} \)[/tex]
- Magnetic field, [tex]\( \vec{B} = 4 \hat{i} + 3 \hat{j} + \hat{k} \, \text{T} \)[/tex]
- Electric field, [tex]\( \vec{E} = 3 \hat{i} - \hat{j} - 2 \hat{k} \, \text{V/m} \)[/tex]
To determine the total force on the charged particle, we need to calculate both the electric force ([tex]\(\vec{F}_e\)[/tex]) and the magnetic force ([tex]\(\vec{F}_b\)[/tex]).
The electric force is given by:
[tex]\[ \vec{F}_e = q \vec{E} \][/tex]
The magnetic force is given by:
[tex]\[ \vec{F}_b = q (\vec{v} \times \vec{B}) \][/tex]
The total force is:
[tex]\[ \vec{F} = \vec{F}_e + \vec{F}_b \][/tex]
Given results for the force components:
[tex]\[ F_x = 3.68 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_y = -3.3119999999999996 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_z = -2.208 \times 10^{-17} \, \text{N} \][/tex]
### Part (b): Angle with the [tex]\( +x \)[/tex]-axis
To find the angle the force vector makes with the positive [tex]\( x \)[/tex]-axis, we use the dot product formula between the force vector and the [tex]\( x \)[/tex]-axis unit vector:
[tex]\[ \theta = \cos^{-1} \left( \frac{\vec{F} \cdot \hat{i}}{|\vec{F}|} \right) \][/tex]
Where
[tex]\[ \vec{F} \cdot \hat{i} = F_x \][/tex]
[tex]\[ |\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} \][/tex]
Given result for the angle:
[tex]\[ \theta \approx 47.24660675818882^\circ \][/tex]
### Part (c): Electric field for zero net force
For the total force on the particle to be zero, the electric force must exactly cancel out the magnetic force:
[tex]\[ \vec{F}_e = -\vec{F}_b \][/tex]
[tex]\[ q \vec{E}_{\text{req}} = -q (\vec{v} \times \vec{B}) \][/tex]
[tex]\[ \vec{E}_{\text{req}} = - (\vec{v} \times \vec{B}) / q \][/tex]
Given results for the required electric field components:
[tex]\[ E_x = -7.0 \, \text{V/m} \][/tex]
[tex]\[ E_y = 8.0 \, \text{V/m} \][/tex]
[tex]\[ E_z = 4.0 \, \text{V/m} \][/tex]
### Summary of Answers
(a) The components of the total force are:
[tex]\[ F_x = 3.68 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_y = -3.3119999999999996 \times 10^{-17} \, \text{N} \][/tex]
[tex]\[ F_z = -2.208 \times 10^{-17} \, \text{N} \][/tex]
(b) The angle the force vector makes with the positive [tex]\( x \)[/tex]-axis is approximately [tex]\( 47.25^\circ \)[/tex] counterclockwise from the [tex]\( +x \)[/tex]-axis.
(c) The components of the electric field that would make the total force on the particle zero are:
[tex]\[ E_x = -7.0 \, \text{V/m} \][/tex]
[tex]\[ E_y = 8.0 \, \text{V/m} \][/tex]
[tex]\[ E_z = 4.0 \, \text{V/m} \][/tex]
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