Question
A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and there is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength?
a
0.60 T
b
6.7 ´ 103 T
c
1.8 ´ 103 T
d
15 T
Question 18 (1 point)
Consider two long, straight, parallel wires, each carrying a current I. If the currents move in opposite directions,
a
the two wires will exert a torque on each other.
b
the two wires will attract each other.
c
the two wires will repel each other.
d
neither wire will exert a force on the other.
Question 19 (1 point)
Consider two long, straight, parallel wires, each carrying a current I. If the currents move in the same direction,
a
neither wire will exert a force on the other.
b
the two wires will repel each other.
c
the two wires will exert a torque on each other.
d
the two wires will attract each other.
Question 20 (1 point)
A current-carrying conductor in and perpendicular to a magnetic field experiences a force that is
a
parallel to the current.
b
inversely proportional to the velocity
c
inversely proportional to the potential difference.
d
perpendicular to the current.
a
0.60 T
b
6.7 ´ 103 T
c
1.8 ´ 103 T
d
15 T
Question 18 (1 point)
Consider two long, straight, parallel wires, each carrying a current I. If the currents move in opposite directions,
a
the two wires will exert a torque on each other.
b
the two wires will attract each other.
c
the two wires will repel each other.
d
neither wire will exert a force on the other.
Question 19 (1 point)
Consider two long, straight, parallel wires, each carrying a current I. If the currents move in the same direction,
a
neither wire will exert a force on the other.
b
the two wires will repel each other.
c
the two wires will exert a torque on each other.
d
the two wires will attract each other.
Question 20 (1 point)
A current-carrying conductor in and perpendicular to a magnetic field experiences a force that is
a
parallel to the current.
b
inversely proportional to the velocity
c
inversely proportional to the potential difference.
d
perpendicular to the current.
Answers
GPT-4o mini
Let's go through each of your questions one by one.
**Question 17: Magnetic Field Strength**
To find the magnetic field strength (B), we can use the formula for the force (F) acting on a current-carrying wire in a magnetic field:
\[
F = B \cdot I \cdot L
\]
Where:
- \( F \) is the force in Newtons (N),
- \( B \) is the magnetic field strength in Tesla (T),
- \( I \) is the current in Amperes (A),
- \( L \) is the length of the wire in meters (m).
Plugging in the values from the question:
- \( F = 3.0 \, \text{N} \)
- \( I = 10.0 \, \text{A} \)
- \( L = 0.50 \, \text{m} \)
We can rearrange the formula to solve for \( B \):
\[
B = \frac{F}{I \cdot L}
\]
Substituting the known values:
\[
B = \frac{3.0 \, \text{N}}{10.0 \, \text{A} \cdot 0.50 \, \text{m}} = \frac{3.0}{5.0} = 0.60 \, \text{T}
\]
So the answer is:
**a) 0.60 T**
---
**Question 18: Forces Between Parallel Wires with Opposite Currents**
When two long, straight, parallel wires carry current in opposite directions, they will exert a **repulsive** force on each other. Therefore, the correct answer is:
**c) the two wires will repel each other.**
---
**Question 19: Forces Between Parallel Wires with Same Currents**
When two long, straight, parallel wires carry current in the same direction, they will exert an **attractive** force on each other. Therefore, the correct answer is:
**d) the two wires will attract each other.**
---
**Question 20: Direction of Force on a Current-Carrying Conductor**
A current-carrying conductor in a magnetic field experiences a force that is **perpendicular** to both the direction of the current and the direction of the magnetic field. Therefore, the correct answer is:
**d) perpendicular to the current.**
---
To summarize, the answers are:
17. a) 0.60 T
18. c) the two wires will repel each other.
19. d) the two wires will attract each other.
20. d) perpendicular to the current.
**Question 17: Magnetic Field Strength**
To find the magnetic field strength (B), we can use the formula for the force (F) acting on a current-carrying wire in a magnetic field:
\[
F = B \cdot I \cdot L
\]
Where:
- \( F \) is the force in Newtons (N),
- \( B \) is the magnetic field strength in Tesla (T),
- \( I \) is the current in Amperes (A),
- \( L \) is the length of the wire in meters (m).
Plugging in the values from the question:
- \( F = 3.0 \, \text{N} \)
- \( I = 10.0 \, \text{A} \)
- \( L = 0.50 \, \text{m} \)
We can rearrange the formula to solve for \( B \):
\[
B = \frac{F}{I \cdot L}
\]
Substituting the known values:
\[
B = \frac{3.0 \, \text{N}}{10.0 \, \text{A} \cdot 0.50 \, \text{m}} = \frac{3.0}{5.0} = 0.60 \, \text{T}
\]
So the answer is:
**a) 0.60 T**
---
**Question 18: Forces Between Parallel Wires with Opposite Currents**
When two long, straight, parallel wires carry current in opposite directions, they will exert a **repulsive** force on each other. Therefore, the correct answer is:
**c) the two wires will repel each other.**
---
**Question 19: Forces Between Parallel Wires with Same Currents**
When two long, straight, parallel wires carry current in the same direction, they will exert an **attractive** force on each other. Therefore, the correct answer is:
**d) the two wires will attract each other.**
---
**Question 20: Direction of Force on a Current-Carrying Conductor**
A current-carrying conductor in a magnetic field experiences a force that is **perpendicular** to both the direction of the current and the direction of the magnetic field. Therefore, the correct answer is:
**d) perpendicular to the current.**
---
To summarize, the answers are:
17. a) 0.60 T
18. c) the two wires will repel each other.
19. d) the two wires will attract each other.
20. d) perpendicular to the current.