A 170-turn solenoid having a length of 25 cm and a diameter of 7.9 cm carries a current of 0.25 A. Calculate the magnitude of the magnetic field inside the solenoid.
A 170-turn solenoid having a length of 25 cm and a diameter of 7.9 cm carries a current of 0.25 A. Calculate the magnitude of the magnetic field inside the solenoid.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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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![**Problem: Calculating the Magnetic Field Inside a Solenoid**
A 170-turn solenoid having a length of 25 cm and a diameter of 7.9 cm carries a current of 0.25 A. Calculate the magnitude of the magnetic field inside the solenoid.
**Solution:**
- **Number of turns (N):** 170
- **Length of solenoid (L):** 25 cm (0.25 meters)
- **Diameter of solenoid:** 7.9 cm (not necessary for this calculation)
- **Current (I):** 0.25 A
Utilize the formula for the magnetic field inside the solenoid (B):
\[ B = \mu_0 \left( \frac{N}{L} \right) I \]
Where:
- \( \mu_0 \) is the permeability of free space ( \( 4 \pi \times 10^{-7} \, T \cdot m/A \) )
**Calculation:**
1. Calculate the term \( \left( \frac{N}{L} \right) \):
\[
\frac{N}{L} = \frac{170 \, \text{turns}}{0.25 \, \text{meters}} = 680 \, \text{turns/m}
\]
2. Calculate the magnetic field (B):
\[
B = (4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \times (680 \, \text{turns/m}) \times (0.25 \, \text{A})
\]
\[
B = 4 \pi \times 10^{-7} \times 680 \times 0.25 \, \text{T}
\]
\[
B = (4 \pi \times 10^{-7}) \times 170 \, \text{T}
\]
\[
B \approx 2.136 \times 10^{-4} \, \text{T}
\]
Therefore, the magnitude of the magnetic field inside the solenoid is approximately \( 2.136 \times 10^{-4} \) Tesla (T).
**Input Representation:**
- **Number:** [Input Box]
- **Units:** [Dropdown Menu]
This input allows](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf1ef7c6-a145-4dbc-a50b-84dc7d2e578d%2F814f2654-413f-4cc9-9117-b0d77c7b33f7%2Fzatjkor_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Calculating the Magnetic Field Inside a Solenoid**
A 170-turn solenoid having a length of 25 cm and a diameter of 7.9 cm carries a current of 0.25 A. Calculate the magnitude of the magnetic field inside the solenoid.
**Solution:**
- **Number of turns (N):** 170
- **Length of solenoid (L):** 25 cm (0.25 meters)
- **Diameter of solenoid:** 7.9 cm (not necessary for this calculation)
- **Current (I):** 0.25 A
Utilize the formula for the magnetic field inside the solenoid (B):
\[ B = \mu_0 \left( \frac{N}{L} \right) I \]
Where:
- \( \mu_0 \) is the permeability of free space ( \( 4 \pi \times 10^{-7} \, T \cdot m/A \) )
**Calculation:**
1. Calculate the term \( \left( \frac{N}{L} \right) \):
\[
\frac{N}{L} = \frac{170 \, \text{turns}}{0.25 \, \text{meters}} = 680 \, \text{turns/m}
\]
2. Calculate the magnetic field (B):
\[
B = (4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \times (680 \, \text{turns/m}) \times (0.25 \, \text{A})
\]
\[
B = 4 \pi \times 10^{-7} \times 680 \times 0.25 \, \text{T}
\]
\[
B = (4 \pi \times 10^{-7}) \times 170 \, \text{T}
\]
\[
B \approx 2.136 \times 10^{-4} \, \text{T}
\]
Therefore, the magnitude of the magnetic field inside the solenoid is approximately \( 2.136 \times 10^{-4} \) Tesla (T).
**Input Representation:**
- **Number:** [Input Box]
- **Units:** [Dropdown Menu]
This input allows
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