A thin, circular coil with 13 turns that has a radius of 16.7 cm carries a current of 2.45 A. What is the strength B of the magnetic field at a point on the coil's axis that is 30.7 cm from the center of the coil? B = T
A thin, circular coil with 13 turns that has a radius of 16.7 cm carries a current of 2.45 A. What is the strength B of the magnetic field at a point on the coil's axis that is 30.7 cm from the center of the coil? B = T
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![**Magnetic Field of a Circular Coil**
Consider a thin, circular coil with 13 turns that has a radius of 16.7 cm and carries a current of 2.45 A. To determine the strength \( B \) of the magnetic field at a point on the coil's axis that is 30.7 cm from the center of the coil, use the following setup and calculations.
### Problem Statement:
A thin, circular coil with 13 turns that has a radius of 16.7 cm carries a current of 2.45 A. What is the strength \( B \) of the magnetic field at a point on the coil's axis that is 30.7 cm from the center of the coil?
### Parameters Given:
- Number of turns, \( N \) = 13
- Radius of the coil, \( R \) = 16.7 cm = 0.167 m
- Current, \( I \) = 2.45 A
- Distance from the center of the coil along its axis, \( z \) = 30.7 cm = 0.307 m
### Formula:
The magnetic field \( B \) on the axis of a circular coil at a distance \( z \) from the center can be calculated using the formula:
\[ B = \frac{(\mu_0 N I R^2)}{2(R^2 + z^2)^{3/2}} \]
where:
- \( \mu_0 \) is the permeability of free space (\( \mu_0 \approx 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \))
- \( N \) is the number of turns
- \( I \) is the current
- \( R \) is the radius of the coil
- \( z \) is the distance from the center of the coil along its axis
### Calculation:
Substitute the given values into the formula to find \( B \).
### Solution Box:
\[
B = \boxed{\text{T}}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37e1db01-06d0-4e20-aae1-26f96b6814c3%2F6c637d4c-bcb6-4a59-b83f-dec1e0f6aefc%2Fp9g8dgd_processed.png&w=3840&q=75)
Transcribed Image Text:**Magnetic Field of a Circular Coil**
Consider a thin, circular coil with 13 turns that has a radius of 16.7 cm and carries a current of 2.45 A. To determine the strength \( B \) of the magnetic field at a point on the coil's axis that is 30.7 cm from the center of the coil, use the following setup and calculations.
### Problem Statement:
A thin, circular coil with 13 turns that has a radius of 16.7 cm carries a current of 2.45 A. What is the strength \( B \) of the magnetic field at a point on the coil's axis that is 30.7 cm from the center of the coil?
### Parameters Given:
- Number of turns, \( N \) = 13
- Radius of the coil, \( R \) = 16.7 cm = 0.167 m
- Current, \( I \) = 2.45 A
- Distance from the center of the coil along its axis, \( z \) = 30.7 cm = 0.307 m
### Formula:
The magnetic field \( B \) on the axis of a circular coil at a distance \( z \) from the center can be calculated using the formula:
\[ B = \frac{(\mu_0 N I R^2)}{2(R^2 + z^2)^{3/2}} \]
where:
- \( \mu_0 \) is the permeability of free space (\( \mu_0 \approx 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \))
- \( N \) is the number of turns
- \( I \) is the current
- \( R \) is the radius of the coil
- \( z \) is the distance from the center of the coil along its axis
### Calculation:
Substitute the given values into the formula to find \( B \).
### Solution Box:
\[
B = \boxed{\text{T}}
\]
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