Question
1.
A proton moves with an unknown velocity through a magnetic field of 3.45 x 10-3 T that points directly north. The proton experiences a force of 2.40 x 10-15 N directly east. What direction is the proton moving?
into the page
out of the page
west
south
2. A proton moves with an unknown velocity through a magnetic field of 3.45 x 10-3 T that points directly north. The proton experiences a force of 2.40 x 10-15 N directly east. What is the magnitude of the velocity? (charge of a proton is +1.6 x 10-19 C)
4.35 x 107 m/s
4.35 x 106 m/s
6.35 x 108 m/s
8.85 x 106 m/s
3. A particle of charge 2.4 x 10-18 C is stationary in a magnetic field of 3.20 T. What is the electric force on the particle caused by the magnetic field?
7.50 x 10-19 N
8.62 x 10-20 N
0 N
7.68 x 10-18 N
4. What was Andre-Marie Ampere known for?
the compass
circuitry
electrodynamics
electromagnetic induction
5. A charged particle moves in a circle in a magnetic field. What must be true about that particle?
the charged particle is moving at an angle to the magnetic field
the charged particle is moving parallel to the magnetic field
the charged particle is moving outside of a magnetic field
the charged particle is moving perpendicularly to a magnetic field
6. An electron moving straight down (into the page) at a speed of 4.82 x 107 m/s experiences a force of 3.07 x 10-12 N directly east. What magnitude of the magnetic field? (The charge of an electron is -1.6 x 10-19 C)
0.148 T
0.39 T
1.48 x 10-4 T
3.26 T
7. Which of the following are true? Check all that apply.
If the two current-carrying wires are placed parallel to one another and the current is moving in opposite directions, the force between them will be attractive.
A current-carrying wire produces a magnetic field around it that moves in a direction given by the right-hand rule for a current-carrying conductor.
If you place a current-carrying wire in a magnetic field, the wire will experience a magnetic force produced by that magnetic field.
If the two current-carrying wires are placed parallel to one another and the current is moving in the same direction, the force between them will be attractive.
8. A current-carrying wire placed in a magnetic field will be deflected by a force that is proportional to: (check all that apply)
the strength of the magnetic field
the type of wire moving in the magnetic field
the length of wire in the magnetic field
the current flowing through the wire
9. An electron moving straight down (into the page) at a speed of 2.75 x 107 m/s experiences a force of 6.07 x 10-12 N directly east. What direction is the magnetic field pointing?
out of the page
west
south
north
10. A proton moving east at 1.30 x 105 m/s moves through a magnetic field of 4.98 x 10-5 T to the north. What is the magnitude of the force that the proton experiences? (charge of a proton is +1.6 x 10-19 C)
1.04 x 10-12 N
4.05 x 10-12 N
4.05 x 10-18 N
1.04 x 10-18 N
A proton moves with an unknown velocity through a magnetic field of 3.45 x 10-3 T that points directly north. The proton experiences a force of 2.40 x 10-15 N directly east. What direction is the proton moving?
into the page
out of the page
west
south
2. A proton moves with an unknown velocity through a magnetic field of 3.45 x 10-3 T that points directly north. The proton experiences a force of 2.40 x 10-15 N directly east. What is the magnitude of the velocity? (charge of a proton is +1.6 x 10-19 C)
4.35 x 107 m/s
4.35 x 106 m/s
6.35 x 108 m/s
8.85 x 106 m/s
3. A particle of charge 2.4 x 10-18 C is stationary in a magnetic field of 3.20 T. What is the electric force on the particle caused by the magnetic field?
7.50 x 10-19 N
8.62 x 10-20 N
0 N
7.68 x 10-18 N
4. What was Andre-Marie Ampere known for?
the compass
circuitry
electrodynamics
electromagnetic induction
5. A charged particle moves in a circle in a magnetic field. What must be true about that particle?
the charged particle is moving at an angle to the magnetic field
the charged particle is moving parallel to the magnetic field
the charged particle is moving outside of a magnetic field
the charged particle is moving perpendicularly to a magnetic field
6. An electron moving straight down (into the page) at a speed of 4.82 x 107 m/s experiences a force of 3.07 x 10-12 N directly east. What magnitude of the magnetic field? (The charge of an electron is -1.6 x 10-19 C)
0.148 T
0.39 T
1.48 x 10-4 T
3.26 T
7. Which of the following are true? Check all that apply.
If the two current-carrying wires are placed parallel to one another and the current is moving in opposite directions, the force between them will be attractive.
A current-carrying wire produces a magnetic field around it that moves in a direction given by the right-hand rule for a current-carrying conductor.
If you place a current-carrying wire in a magnetic field, the wire will experience a magnetic force produced by that magnetic field.
If the two current-carrying wires are placed parallel to one another and the current is moving in the same direction, the force between them will be attractive.
8. A current-carrying wire placed in a magnetic field will be deflected by a force that is proportional to: (check all that apply)
the strength of the magnetic field
the type of wire moving in the magnetic field
the length of wire in the magnetic field
the current flowing through the wire
9. An electron moving straight down (into the page) at a speed of 2.75 x 107 m/s experiences a force of 6.07 x 10-12 N directly east. What direction is the magnetic field pointing?
out of the page
west
south
north
10. A proton moving east at 1.30 x 105 m/s moves through a magnetic field of 4.98 x 10-5 T to the north. What is the magnitude of the force that the proton experiences? (charge of a proton is +1.6 x 10-19 C)
1.04 x 10-12 N
4.05 x 10-12 N
4.05 x 10-18 N
1.04 x 10-18 N
Answers
GPT-4o mini
Let's break this down question by question:
### 1. Direction of the Proton
The force on a charged particle moving in a magnetic field is given by the right-hand rule. Point your fingers in the direction of the velocity (unknown in this case), curl them towards the magnetic field (north), and your thumb will point in the direction of the force (east). Thus, to have a force pointing east with a magnetic field pointing north, the proton must be moving **into the page**.
**Answer: into the page**
---
### 2. Magnitude of Velocity of the Proton
The force on a charged particle in a magnetic field is given by \( F = q v B \sin(\theta) \), where \( \theta \) is the angle between the velocity and the magnetic field.
For the proton:
- Force \( F = 2.40 \times 10^{-15} \, \text{N} \)
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field \( B = 3.45 \times 10^{-3} \, \text{T} \)
- The angle \( \theta = 90^\circ \) (since it is into the page moving perpendicular to the magnetic field)
\[
F = q v B
\]
\[
2.40 \times 10^{-15} = (1.6 \times 10^{-19}) v (3.45 \times 10^{-3})
\]
Solving for \( v \):
\[
v = \frac{2.40 \times 10^{-15}}{(1.6 \times 10^{-19}) (3.45 \times 10^{-3})}
\]
\[
v \approx 4.35 \times 10^{7} \, \text{m/s}
\]
**Answer: 4.35 x 10^7 m/s**
---
### 3. Electric Force on the Stationary Particle
A stationary charged particle does not experience a magnetic force because the velocity (v) is zero.
**Answer: 0 N**
---
### 4. Ampere's Contribution
Andre-Marie Ampere is best known for his work in **electrodynamics** and the formulation of Ampere's Law.
**Answer: electrodynamics**
---
### 5. Charged Particle in a Magnetic Field
For a charged particle to move in a circle in a magnetic field, it must be moving **perpendicularly to the magnetic field**. This ensures that the magnetic force acts as a centripetal force.
**Answer: the charged particle is moving perpendicularly to a magnetic field**
---
### 6. Magnitude of the Magnetic Field
Using the formula \( F = q v B \):
Given:
- Force \( F = 3.07 \times 10^{-12} \, \text{N} \)
- Charge \( q = -1.6 \times 10^{-19} \, \text{C} \) (magnitude will be used)
- Velocity \( v = 4.82 \times 10^{7} \, \text{m/s} \)
\[
B = \frac{F}{qv} = \frac{3.07 \times 10^{-12}}{(1.6 \times 10^{-19}) (4.82 \times 10^{7})}
\]
Calculating will yield:
\[
B \approx 0.39 \, \text{T}
\]
**Answer: 0.39 T**
---
### 7. True Statements
1. If the two current-carrying wires are placed parallel to one another and the current is moving in opposite directions, the force between them will be attractive. **(False)**
2. A current-carrying wire produces a magnetic field around it that moves in a direction given by the right-hand rule for a current-carrying conductor. **(True)**
3. If you place a current-carrying wire in a magnetic field, the wire will experience a magnetic force produced by that magnetic field. **(True)**
4. If the two current-carrying wires are placed parallel to one another and the current is moving in the same direction, the force between them will be attractive. **(True)**
**Answer: True statements are: (2), (3), (4)**
---
### 8. Factors of Force on Wire in Magnetic Field
The force on a current-carrying wire depends on:
1. The strength of the magnetic field
2. The length of wire in the magnetic field
3. The current flowing through the wire
**Answer: the strength of the magnetic field, the length of wire in the magnetic field, the current flowing through the wire**
---
### 9. Direction of the Magnetic Field
Using the right-hand rule: The electron is moving straight down (into the page), and the force experienced is to the east. According to the right-hand rule, if the thumb points in the direction of the velocity (down), and the fingers point in the direction of the magnetic field (that we need to determine), the palm will push east (the direction of force) which means the magnetic field has to point **out of the page**.
**Answer: out of the page**
---
### 10. Magnitude of Force on Proton
Calculating the force experienced by the proton using \( F = q v B \):
Given:
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Velocity \( v = 1.30 \times 10^{5} \, \text{m/s} \)
- Magnetic field \( B = 4.98 \times 10^{-5} \, \text{T} \)
\[
F = q v B = (1.6 \times 10^{-19}) (1.30 \times 10^{5}) (4.98 \times 10^{-5})
\]
Calculating, you would find:
\[
F \approx 1.04 \times 10^{-18} \, \text{N}
\]
**Answer: 1.04 x 10^{-18} N**
### 1. Direction of the Proton
The force on a charged particle moving in a magnetic field is given by the right-hand rule. Point your fingers in the direction of the velocity (unknown in this case), curl them towards the magnetic field (north), and your thumb will point in the direction of the force (east). Thus, to have a force pointing east with a magnetic field pointing north, the proton must be moving **into the page**.
**Answer: into the page**
---
### 2. Magnitude of Velocity of the Proton
The force on a charged particle in a magnetic field is given by \( F = q v B \sin(\theta) \), where \( \theta \) is the angle between the velocity and the magnetic field.
For the proton:
- Force \( F = 2.40 \times 10^{-15} \, \text{N} \)
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field \( B = 3.45 \times 10^{-3} \, \text{T} \)
- The angle \( \theta = 90^\circ \) (since it is into the page moving perpendicular to the magnetic field)
\[
F = q v B
\]
\[
2.40 \times 10^{-15} = (1.6 \times 10^{-19}) v (3.45 \times 10^{-3})
\]
Solving for \( v \):
\[
v = \frac{2.40 \times 10^{-15}}{(1.6 \times 10^{-19}) (3.45 \times 10^{-3})}
\]
\[
v \approx 4.35 \times 10^{7} \, \text{m/s}
\]
**Answer: 4.35 x 10^7 m/s**
---
### 3. Electric Force on the Stationary Particle
A stationary charged particle does not experience a magnetic force because the velocity (v) is zero.
**Answer: 0 N**
---
### 4. Ampere's Contribution
Andre-Marie Ampere is best known for his work in **electrodynamics** and the formulation of Ampere's Law.
**Answer: electrodynamics**
---
### 5. Charged Particle in a Magnetic Field
For a charged particle to move in a circle in a magnetic field, it must be moving **perpendicularly to the magnetic field**. This ensures that the magnetic force acts as a centripetal force.
**Answer: the charged particle is moving perpendicularly to a magnetic field**
---
### 6. Magnitude of the Magnetic Field
Using the formula \( F = q v B \):
Given:
- Force \( F = 3.07 \times 10^{-12} \, \text{N} \)
- Charge \( q = -1.6 \times 10^{-19} \, \text{C} \) (magnitude will be used)
- Velocity \( v = 4.82 \times 10^{7} \, \text{m/s} \)
\[
B = \frac{F}{qv} = \frac{3.07 \times 10^{-12}}{(1.6 \times 10^{-19}) (4.82 \times 10^{7})}
\]
Calculating will yield:
\[
B \approx 0.39 \, \text{T}
\]
**Answer: 0.39 T**
---
### 7. True Statements
1. If the two current-carrying wires are placed parallel to one another and the current is moving in opposite directions, the force between them will be attractive. **(False)**
2. A current-carrying wire produces a magnetic field around it that moves in a direction given by the right-hand rule for a current-carrying conductor. **(True)**
3. If you place a current-carrying wire in a magnetic field, the wire will experience a magnetic force produced by that magnetic field. **(True)**
4. If the two current-carrying wires are placed parallel to one another and the current is moving in the same direction, the force between them will be attractive. **(True)**
**Answer: True statements are: (2), (3), (4)**
---
### 8. Factors of Force on Wire in Magnetic Field
The force on a current-carrying wire depends on:
1. The strength of the magnetic field
2. The length of wire in the magnetic field
3. The current flowing through the wire
**Answer: the strength of the magnetic field, the length of wire in the magnetic field, the current flowing through the wire**
---
### 9. Direction of the Magnetic Field
Using the right-hand rule: The electron is moving straight down (into the page), and the force experienced is to the east. According to the right-hand rule, if the thumb points in the direction of the velocity (down), and the fingers point in the direction of the magnetic field (that we need to determine), the palm will push east (the direction of force) which means the magnetic field has to point **out of the page**.
**Answer: out of the page**
---
### 10. Magnitude of Force on Proton
Calculating the force experienced by the proton using \( F = q v B \):
Given:
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Velocity \( v = 1.30 \times 10^{5} \, \text{m/s} \)
- Magnetic field \( B = 4.98 \times 10^{-5} \, \text{T} \)
\[
F = q v B = (1.6 \times 10^{-19}) (1.30 \times 10^{5}) (4.98 \times 10^{-5})
\]
Calculating, you would find:
\[
F \approx 1.04 \times 10^{-18} \, \text{N}
\]
**Answer: 1.04 x 10^{-18} N**