Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
