Problem 42 - University Physics with Modern Physics

Question

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.40 \(\mathrm{m}\) long and 20.0 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Step-by-Step Solution

Answer

You need a current of approximately 16.67 A.

Step 1: Understand the Problem

We need to find the current required to create a magnetic field of 0.150 T in a solenoid. The solenoid has 4000 turns, a length of 1.40 m, and a diameter of 20.0 cm.

Step 2: Recall the Formula for Magnetic Field in a Solenoid

The magnetic field inside a solenoid is given by the formula: \( B = \mu_{0} \cdot n \cdot I \)where \( B \) is the magnetic field (0.150 T), \( \mu_{0} \) is the permeability of free space \( (4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A}) \), \( n \) is the number of turns per unit length, and \( I \) is the current.

Step 3: Calculate the Number of Turns per Unit Length

The number of turns per meter \( n \) is equal to the total number of turns divided by the length of the solenoid: \( n = \frac{4000}{1.40 \; \text{m}} = 2857.14 \; \text{turns/m} \).

Step 4: Rearrange the Formula to Solve for Current \( I \)

Rearrange the formula to solve for \( I \): \( I = \frac{B}{\mu_{0} \cdot n} \).

Step 5: Substitute Values into the Formula

Substitute the known values into the equation: \( I = \frac{0.150 \; \text{T}}{(4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A}) \cdot 2857.14 \; \text{turns/m}} \).

Step 6: Calculate the Current \( I \)

Perform the calculation: \( I = \frac{0.150}{(4\pi \times 10^{-7}) \times 2857.14} \approx 16.67 \; \text{A} \).

Key Concepts

Solenoid Design

A solenoid is a simple, yet powerful, component commonly used in electronics to create magnetic fields. When designing a solenoid, it's crucial to consider its physical dimensions and the number of turns of wire it has. In practical applications, the solenoid's design determines the strength and uniformity of the magnetic field it can produce.

In the given exercise, the solenoid must be 1.40 meters long and have a diameter of 20.0 centimeters, which are its length and width. This specific design impacts how the magnetic field forms within the solenoid. The longer the solenoid, the more uniform its magnetic field tends to be along the majority of its length. Meanwhile, the diameter primarily influences the physical space it occupies and not directly the magnetic field strength.

The number of wire turns is another critical factor. With 4000 turns, this particular solenoid wraps a significant amount of wire within its confines, which helps intensify the magnetic field. The tight coil of wire increases the magnetic field internally, allowing for efficient electromagnetic force generation. How the wire is wound also matters, as it needs to be orderly to maximize the field produced at the center. This feature ensures that solenoids are designed not just for capacity but also for functionality.

Magnetic Field Calculation

To calculate the magnetic field in a solenoid, we use the formula:\[ B = \mu_{0} \cdot n \cdot I \]where:

• \(B\) is the magnetic field
• \(\mu_{0}\) is the permeability of free space, a constant valued at \(4\pi \times 10^{-7} \; \text{T} \cdot \text{m/A}\)
• \(n\) represents the number of turns per unit length of the solenoid
• \(I\) is the current passing through the coil

To find the value of the magnetic field at the center of the solenoid, knowing the number of coils and the length is vital. As calculated in the exercise, the number of turns per meter is \( n = \frac{4000}{1.40 \; \text{m}} = 2857.14 \; \text{turns/m} \).

The formula illustrates that the magnetic field inside a solenoid is directly proportional to the current (\(I\)) and the number of turns per unit length (\(n\)). This means that increasing the current or the coil density enhances the magnetic field strength.

Solenoids are uniquely advantageous in generating controlled magnetic fields because their design channels the field lines efficiently within their hollow interior, making them essential in applications like MRI machines and electromagnets.

Current in Solenoids

The current running through the wire coil of a solenoid is a decisive factor in the magnetic field it generates. In the formula for the magnetic field of a solenoid, the current \(I\) can be determined by rearranging the equation:\[ I = \frac{B}{\mu_{0} \cdot n} \]By substituting the known values for a solenoid designed to produce a magnetic field of 0.150 Tesla, we can calculate the needed current. Plugging in the values, we find:\[ I = \frac{0.150 \; \text{T}}{(4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A}) \cdot 2857.14 \; \text{turns/m}} \approx 16.67 \; \text{A} \]

Thus, a current of approximately 16.67 amperes is required. This is a significant amount of current, which highlights the need for robust wire and possibly a cooling mechanism, as high currents can generate heat.

The calculation of current helps ensure the solenoid operates efficiently and safely, allowing it to perform its intended function without risking damage or failure due to overheating. Understanding how to manipulate current and manage electrical efficiency is vital for technicians working with solenoids, as it directly impacts performance and durability.