Problem 9 - Applied Physics

Question

A small solenoid is \(0.150 \mathrm{~m}\) in length and \(0.0150 \mathrm{~m}\) in diameter, and has \(60 \overline{0}\) turns of wire. What current is required to produce a magnetic field of \(1.25 \times 10^{-3} \mathrm{~T}\) at the center of the solenoid?

Step-by-Step Solution

Answer

The required current is approximately 248.7 mA.

Step 1: Understand the Parameters of the Solenoid

The parameters given in the problem are: the length of the solenoid \(L = 0.150 \, \text{m}\), the diameter \(d = 0.0150 \, \text{m}\), the number of turns \(N = 600\), and the magnetic field \(B = 1.25 \times 10^{-3} \, \text{T}\). The goal is to find the current \(I\) needed to produce this magnetic field.

Step 2: Use the Formula for the Magnetic Field Inside a Solenoid

The magnetic field inside a long solenoid is given by the formula: \( B = \mu_0 \frac{N}{L} I \). Here, \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7} \, \text{T} \, \text{m/A}\). We aim to solve this equation for \(I\).

Step 3: Rearrange the Formula to Solve for Current

Rearrange the formula to find the current by dividing both sides by \(\mu_0 \frac{N}{L}\): \[ I = \frac{B L}{\mu_0 N} \]

Step 4: Substitute the Known Values into the Formula

Substitute \(B = 1.25 \times 10^{-3} \, \text{T}\), \(L = 0.150 \, \text{m}\), \(N = 600\), and \(\mu_0 = 4\pi \times 10^{-7} \, \text{T} \, \text{m/A}\) into the equation: \[ I = \frac{(1.25 \times 10^{-3}) \times 0.150}{4\pi \times 10^{-7} \times 600} \]

Step 5: Calculate to Find the Current

Perform the calculation: \( I = \frac{1.25 \times 0.150}{4\pi \times 600 \times 10^{-7}} = \frac{0.1875}{7.54 \times 10^{-4}} \approx 248.7\) A. Thus, the current required is approximately \(0.2487 \, \text{A}\) or 248.7 mA.

Key Concepts

Solenoid

A solenoid is a type of coil that is commonly used in electromagnetism to create a uniform magnetic field in its interior. It is essentially a long, cylindrical coil of wire, and when an electric current passes through it, a magnetic field is produced. The strength of this magnetic field depends largely on:

• The number of turns of wire, which directly affects the magnetic field intensity.
• The length of the solenoid, as a longer solenoid can distribute the field more evenly.
• The current flowing through the wire, which amplifies the magnetic field as it increases.

For the given solenoid, a length of 0.150 meters and 600 turns of wire implies that this solenoid is designed to create a consistent field along its length, provided a current is applied. This uniformity is why solenoids are so useful in applications like transformers and electromagnets.
Understanding the structure of a solenoid can help appreciate how the design effectively channels magnetic lines of force through its center to create a field. The arrangement of the turns, closely packed, strengthens the magnetic lines of force, making solenoids powerful electromagnetic tools.

Magnetic Field

The magnetic field in the center of a solenoid is an important feature of this component as it dictates many practical uses. The magnetic field (\( B \)) within a solenoid is uniform, meaning its field strength is the same at every point inside the coil. The formula to find the magnetic field inside a solenoid is \( B = \mu_0 \frac{N}{L} I \), where:

• \( \mu_0 \) represents the permeability of free space, a constant value.
• \( N \) is the number of turns.
• \( L \) is the length of the solenoid.
• \( I \) is the current through the solenoid.

The uniform magnetic field generated by the solenoid is pivotal for its ability to concentrate magnetic force in applications like inductors and MRI machines.
In the given exercise, the magnetic field you want to achieve inside the solenoid is \( 1.25 \times 10^{-3} \) Tesla, which requires precise calculation of current to ensure the solenoid works as intended. This understanding of the magnetic field helps in engineering solutions that necessitate precise magnetic properties.

Current Calculation

Calculating the current required in a solenoid to produce a desired magnetic field involves rearranging the solenoid magnetic field formula. Solving for current \( I \), the equation becomes \( I = \frac{B L}{\mu_0 N} \). This calculation takes into account:

• The desired magnetic field strength (\( B \)): This is what you want the solenoid to produce.
• The solenoid's physical parameters (\( N \) and \( L \)): Representing how the solenoid is built.
• The properties of the material (\( \mu_0 \)): Which tells how well a magnetic field can be induced.

For the problem at hand, inserting our known values into the formula gives the needed current as about 0.2487 Amperes or 248.7 mA. This solution exemplifies the reliance on precise current values to achieve specified magnetic fields, critical in designing electromagnetic devices.
Understanding these calculations ensures effective use of solenoids in technology requiring controlled magnetic fields, such as motors and solenoidal valves.

Permeability

Permeability is a fundamental concept in magnetism that defines how a material responds to a magnetic field and its ability to support the formation of a magnetic field within itself. In the context of a solenoid, the permeability of free space, denoted by \( \mu_0 \), is approximately \( 4\pi \times 10^{-7} \ ext{T m/A} \). This constant is crucial for calculating the magnetic field produced by a solenoid.

• Permeability influences how efficiently a solenoid can produce a magnetic field.
• A higher permeability means a stronger magnetic field can be achieved with less current.
• A solenoid's core material can alter the effective permeability, further enhancing field strength.

In our exercise, \( \mu_0 \) maintains its default constant value because we assume a vacuum or air-filled core, both of which have minimal impact on the base permeability. Understanding permeability helps in designing solenoids that function optimally by manipulating the core material to meet specific performance needs.