The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long permanent magnet is 0.10 T. Part A To produce the same field with a solenoid of the same size, carrying a current of 1.8 A, how many turns of wire would you need? Express your answer to two significant figures. 15. ΑΣΦ ? N = Submit Previous Answers Request Answer turns
The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long permanent magnet is 0.10 T. Part A To produce the same field with a solenoid of the same size, carrying a current of 1.8 A, how many turns of wire would you need? Express your answer to two significant figures. 15. ΑΣΦ ? N = Submit Previous Answers Request Answer turns
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![### Magnetic Field of a Permanent Magnet and Solenoid
#### Context
This content relates to understanding the magnetic field strength produced by a permanent magnet and comparing it to that produced by a solenoid.
#### Problem Statement
The magnetic field strength at the north pole of a permanent magnet, which is 2.0 cm in diameter and 8 cm in length, is 0.10 T.
#### Objective
To produce the same magnetic field strength with a solenoid of the same size, carrying a current of 1.8 A, how many turns of wire would be needed?
#### Instructions
- **Express your answer to two significant figures.**
##### Input Field:
- \( N = \) [__________] turns
##### Actions Available:
- **Submit**
- **Previous Answers**
- **Request Answer**
The user is expected to calculate the number of turns (\( N \)) of the wire required for the solenoid to produce a magnetic field of 0.10 T given a current of 1.8 A.
### Background Information
- **Permanent Magnet**: A magnet that maintains its magnetic properties even in the absence of an external magnetic field.
- **Solenoid**: A coil of wire designed to generate a magnetic field when an electric current passes through it.
- **Magnetic Field (B)**: The magnetic influence produced by electric currents and magnetic materials.
#### Input Parameters:
- Diameter of magnet: 2.0 cm
- Length of magnet: 8 cm
- Magnetic field strength: 0.10 T (tesla)
- Solenoid current: 1.8 A (amperes)
### Formula (For self-calculation)
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 \frac{N}{L} I \]
Where:
- \( B \) is the magnetic field strength (tesla, T)
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T·m/A} \))
- \( N \) is the number of turns of the coil
- \( L \) is the length of the solenoid (meters, m)
- \( I \) is the current through the solenoid (amperes, A)
#### Application
Based on these parameters, someone can calculate the required number of turns \( N \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff89c5e82-2527-49f4-96f0-d8c4ad99581b%2F6a7c184d-7692-4855-bccb-61a3cced9f5d%2F0q5ay6d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Magnetic Field of a Permanent Magnet and Solenoid
#### Context
This content relates to understanding the magnetic field strength produced by a permanent magnet and comparing it to that produced by a solenoid.
#### Problem Statement
The magnetic field strength at the north pole of a permanent magnet, which is 2.0 cm in diameter and 8 cm in length, is 0.10 T.
#### Objective
To produce the same magnetic field strength with a solenoid of the same size, carrying a current of 1.8 A, how many turns of wire would be needed?
#### Instructions
- **Express your answer to two significant figures.**
##### Input Field:
- \( N = \) [__________] turns
##### Actions Available:
- **Submit**
- **Previous Answers**
- **Request Answer**
The user is expected to calculate the number of turns (\( N \)) of the wire required for the solenoid to produce a magnetic field of 0.10 T given a current of 1.8 A.
### Background Information
- **Permanent Magnet**: A magnet that maintains its magnetic properties even in the absence of an external magnetic field.
- **Solenoid**: A coil of wire designed to generate a magnetic field when an electric current passes through it.
- **Magnetic Field (B)**: The magnetic influence produced by electric currents and magnetic materials.
#### Input Parameters:
- Diameter of magnet: 2.0 cm
- Length of magnet: 8 cm
- Magnetic field strength: 0.10 T (tesla)
- Solenoid current: 1.8 A (amperes)
### Formula (For self-calculation)
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 \frac{N}{L} I \]
Where:
- \( B \) is the magnetic field strength (tesla, T)
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T·m/A} \))
- \( N \) is the number of turns of the coil
- \( L \) is the length of the solenoid (meters, m)
- \( I \) is the current through the solenoid (amperes, A)
#### Application
Based on these parameters, someone can calculate the required number of turns \( N \)
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