### Charged Particle in Electric and Magnetic Fields#### Problem Statement:A negatively charged particle (mass \( m = 5.0 \, \text{g} \), charge \( q = -70 \, \mu\text{C} \)) moves horizontally at a constant speed of \( 30 \, \text{km/s} \) in a region where the free-fall gravitational acceleration is \( 9.8 \, \text{m/s}^2 \) downward, the electric field is \( 700 \, \text{N/C} \) upward, and the magnetic field is perpendicular to the velocity of the particle.#### Questions:1. **What are the directions of the electric and magnetic forces on the particle?** - [ ] Both forces are upward - [ ] Electric force downward; magnetic force upward - [ ] Both forces are downward - [x] Electric force upward; magnetic force downward2. **What is the magnitude of the magnetic field in this region?** \( \boxed{} \) \( \text{T} \)#### Detailed Explanation:1. **Forces on the Particle:** - Since the particle is negatively charged, the direction of the electric force will be opposite to the direction of the electric field. Given that the electric field is upward, the electric force on the negatively charged particle will be downward. - The direction of the magnetic force is determined by the right-hand rule. However, for a negatively charged particle, the force direction is opposite to what the right-hand rule would predict for a positive charge. Given that the magnetic field is perpendicular to the velocity, the force will be in the plane perpendicular to both, directed downward due to the negative charge.2. **Magnitude of the Magnetic Field:** - To find the magnitude of the magnetic field (\(B\)), we need to consider the balance of forces acting on the particle which allows it to move at a constant speed. - The gravitational force (\( F_g \)) acts downward: \( F_g = m \cdot g \) - The electric force (\( F_e \)) acts upward: \( F_e = q \cdot E \) - The magnetic force (\( F_B \)) acts downward: \( F_B = q \cdot v \cdot B \)Considering equilibrium in the vertical direction:\[ F_e = F

Answered: Image /qna-images/answer/3589dc4c-a212-416b-b564-b0e7f82c8caa.jpg

A negatively charged particle (m = 5.0 g, q = -70 microC) moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s2 downward, the electric field is 700 N/C upward, and the magnetic field is perpendicular to the velocity of the particle. a) What are the directions of the electric and magnetic forces on the particle? O Both forces are upward O Electric force downward; magnetic force upward O Both forces are downward O Electric force upward; magnetic force downward b) What is the magnitude of the magnetic field in this region?

A negatively charged particle (m = 5.0 g, q = -70 microC) moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s2 downward, the electric field is 700 N/C upward, and the magnetic field is perpendicular to the velocity of the particle. a) What are the directions of the electric and magnetic forces on the particle? O Both forces are upward O Electric force downward; magnetic force upward O Both forces are downward O Electric force upward; magnetic force downward b) What is the magnitude of the magnetic field in this region?

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ISBN:9781305952300

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Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

Question

### Charged Particle in Electric and Magnetic Fields

#### Problem Statement:
A negatively charged particle (mass \( m = 5.0 \, \text{g} \), charge \( q = -70 \, \mu\text{C} \)) moves horizontally at a constant speed of \( 30 \, \text{km/s} \) in a region where the free-fall gravitational acceleration is \( 9.8 \, \text{m/s}^2 \) downward, the electric field is \( 700 \, \text{N/C} \) upward, and the magnetic field is perpendicular to the velocity of the particle.

#### Questions:
1. **What are the directions of the electric and magnetic forces on the particle?**
- [ ] Both forces are upward
- [ ] Electric force downward; magnetic force upward
- [ ] Both forces are downward
- [x] Electric force upward; magnetic force downward

2. **What is the magnitude of the magnetic field in this region?**

\( \boxed{} \) \( \text{T} \)

#### Detailed Explanation:
1. **Forces on the Particle:**
- Since the particle is negatively charged, the direction of the electric force will be opposite to the direction of the electric field. Given that the electric field is upward, the electric force on the negatively charged particle will be downward.
- The direction of the magnetic force is determined by the right-hand rule. However, for a negatively charged particle, the force direction is opposite to what the right-hand rule would predict for a positive charge. Given that the magnetic field is perpendicular to the velocity, the force will be in the plane perpendicular to both, directed downward due to the negative charge.

2. **Magnitude of the Magnetic Field:**
- To find the magnitude of the magnetic field (\(B\)), we need to consider the balance of forces acting on the particle which allows it to move at a constant speed.
- The gravitational force (\( F_g \)) acts downward: \( F_g = m \cdot g \)
- The electric force (\( F_e \)) acts upward: \( F_e = q \cdot E \)
- The magnetic force (\( F_B \)) acts downward: \( F_B = q \cdot v \cdot B \)

Considering equilibrium in the vertical direction:
\[ F_e = F

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