Problem 11 - Applied Physics

Question

A long solenoid has 1000 turns and is \(0.250 \mathrm{~m}\) long. If the wire carries a current of \(2.50 \mathrm{~A}\), what is the strength of the magnetic field at its center?

Step-by-Step Solution

Answer

The magnetic field strength at the center of the solenoid is approximately 0.01257 T.

Step 1: Identify Given Values

We have a solenoid with 1000 turns ( n = 1000 ), length (L) of 0.250 meters, and current (I) of 2.50 amperes.

Step 2: Calculate Number of Turns per Unit Length

The number of turns per unit length n' is given by the formula:\[ n' = \frac{n}{L} \]Substitute the known values:\[ n' = \frac{1000}{0.250} = 4000 \text{ turns/m} \]

Step 3: Apply Solenoid Magnetic Field Formula

The magnetic field (B) inside a long solenoid is given by:\[ B = \mu_0 \, n' \, I \]where \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \) is the permeability of free space.

Step 4: Substitute Values into Formula

Using the values calculated and given, substitute them into the magnetic field formula:\[ B = (4\pi \times 10^{-7} \, \text{T m/A}) \, \times 4000 \, \text{turns/m} \times 2.50 \, \text{A}\]

Step 5: Calculate Magnetic Field Strength

Calculate the magnetic field:\[ B = (4\pi \times 10^{-7}) \, \times 4000 \, \times 2.50 \approx 0.01257 \, \text{T} \]

Key Concepts

Permeability of Free Space

Understanding the concept of permeability of free space, often symbolized as \( \mu_0 \), is vital when dealing with electromagnetic topics, such as solenoids. This constant represents how much magnetic field is supported in a vacuum, serving as a fundamental property in physics. It is quantified as \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
This value allows us to calculate the magnetic field produced in solenoids given other parameters like current and turns per unit length. By using \( \mu_0 \), one relates magnetic quantities in a straightforward manner for calculations.

• Widely used in numerous formulas, such as the magnetic field in a solenoid.
• Helps define magnetic behavior in vacuum conditions.
• Essential for understanding electromagnetism as a whole.

Consider \( \mu_0 \) as the bridge between electrical currents and the magnetic fields they generate, making it invaluable for calculating magnetic field strength in a pristine environment.

Turns per Unit Length

To analyze a solenoid properly, one must grasp the concept of turns per unit length, denoted as \( n' \). It represents how tightly the wire is coiled in the solenoid. The formula used is \( n' = \frac{n}{L} \), where \( n \) indicates the total number of turns and \( L \) is the solenoid's length.
This concept is critical because it directly influences the magnetic field generated: a greater number of turns per unit length results in a stronger magnetic field.

• Higher turns per unit length lead to stronger induced magnetic fields.
• A key factor in tuning the magnetic field strength within solenoids.
• Crucial for various applications, such as electromagnets and transformers.

It's important to visualize turns per unit length as the density of the coil—more density means more power in the magnetic field it produces, making it a pivotal parameter in magnetic field management.

Magnetic Field Strength Calculation

Calculating the magnetic field strength at the center of a solenoid involves an interplay of the solenoid's current, turns per unit length, and the permeability of free space. The magnetic field \( B \) can be calculated using the formula: \( B = \mu_0 \cdot n' \cdot I \), where:

• \( \mu_0 \) is the permeability of free space.
• \( n' \) is the turns per unit length.
• \( I \) is the current flowing through the wire.

Consider each component:- **Permeability of Free Space (\( \mu_0 \))** contributes to the field by defining the medium's support for the magnetic lines.
- **Turns per Unit Length (\( n' \))** indicates how effectively the coil wraps can generate magnetic lines.
- **Current (\( I \))** relates to the source driving the field: stronger current means a stronger field.
Bringing these together enables accurate magnetic field calculations, as seen in the given solution where substituting known values yields a magnetic field, \( B \approx 0.01257 \, \text{T} \). Proper utilization of these variables allows for precise electromagnetic manipulation in various engineering and physics applications.