Question

A 0.90 m-long solenoid has a radius of \(5.0 \mathrm{~mm} .\) When the wire carries a 0.20 - A current, the magnetic field in the solenoid is \(5.0 \mathrm{mT}\). How many turns of wire are there in the solenoid?

Short answer

Answer: There are approximately 179 turns of wire in the solenoid.

Step-by-step solution

Write down the given information

We are given the following information: Length of solenoid, \(L = 0.90 \, \mathrm{m}\) Radius of solenoid, \(r = 5.0 \, \mathrm{mm}\) Current in the wire, \(I = 0.20 \, \mathrm{A}\) Magnetic field inside the solenoid, \(B = 5.0 \, \mathrm{mT}\)

Convert units if necessary

We must ensure that the units of our given quantities are consistent. In this case, we have the radius in millimeters and the magnetic field in milliteslas. We need to convert these to SI units for meters and teslas, respectively. \(r = 5.0 \, \mathrm{mm} = 5.0 \times 10^{-3} \, \mathrm{m}\) \(B = 5.0 \, \mathrm{mT} = 5.0 \times 10^{-3} \, \mathrm{T}\)

Apply the formula for the magnetic field inside a solenoid

The magnetic field inside a solenoid is given by \(B = \mu_0 * n * I\), where \(\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{T\cdot m/A}\) is the permeability of free space. In this case, we need to solve for \(n\), the number of turns per unit length. Rearrange the formula to solve for \(n\): \(n = \frac{B}{\mu_0 * I}\)

Substitute the given values and solve for \(n\)

Now we can substitute our given values and constants into the formula for \(n\): \(n = \frac{5.0 \times 10^{-3} \, \mathrm{T}}{4 \pi \times 10^{-7} \, \mathrm{T\cdot m/A} * 0.20 \, \mathrm{A}}\) Calculate \(n\): \(n \approx 198.94 \, \mathrm{turns/m}\)

Determine the total number of turns in the solenoid

To find the total number of turns in the solenoid, we need to multiply the number of turns per unit length by the solenoid's total length \(L = 0.90 \, \mathrm{m}\): Total turns = \(n * L\) Total turns = \(198.94 \, \mathrm{turns/m} * 0.90 \, \mathrm{m} \approx 179.05 \, \mathrm{turns}\)

Round to the nearest integer

Since the number of turns must be an integer value, we will round our result to the nearest whole number: Total turns \(\approx 179 \, \mathrm{turns}\) So, there are approximately 179 turns of wire in the solenoid.

Key concepts

Magnetic Field Inside a Solenoid

Understanding the behavior of the magnetic field inside a solenoid is significant for grasping basic electromagnetic concepts. A solenoid is a type of electromagnet consisting of a wire wound into a tightly packed helix. The magnetic field within a solenoid is remarkably uniform, making it a fascinating subject of study.

When an electric current passes through the coil, a magnetic field is produced throughout the length of the solenoid, with the field lines running parallel to the axis of the solenoid. This uniformity is a result of the shape and the winding of the wire. Notably, the magnetic field’s strength can be manipulated by changing the current or the number of coils per unit length, which influences the intensity of the magnetic field produced.

The magnetic field inside a solenoid \(B\) can be calculated using the formula \(B = \mu_0 \times n \times I\), where \(\mu_0\) is the permeability of free space, \(n\) is the turns per unit length, and \(I\) is the current running through the solenoid. This relationship is central to solenoid applications in various devices, such as MRI machines or electromagnetic valves.

Permeability of Free Space

The permeability of free space, denoted as \(\mu_0\), is one of the fundamental constants in electromagnetism. It represents how much resistance the vacuum of free space offers against the formation of a magnetic field. The value of \(\mu_0\) is approximately \(4\pi \times 10^{-7} \, \mathrm{T\cdot m/A}\) under classical physics.

When calculating the magnetic field in various environments, permeability plays a crucial role in determining the magnetic field’s strength. In a vacuum environment, \(\mu_0\) provides a base level for the magnetic field calculations, and in materials other than the vacuum, the material's permeability often comes into play.

In the exercise from the textbook, \(\mu_0\) is crucial in the formula used to calculate the solenoid's magnetic field. It acts as a proportionality constant that relates the magnetic field in the solenoid to the number of turns per unit length and the current passing through the solenoid.

Turns Per Unit Length

The concept of turns per unit length \(n\) is an essential factor that determines the strength of the magnetic field in a solenoid. It represents the number of coil turns of the wire that are wrapped in a one-meter length of the solenoid.

The higher the turns per unit length, the stronger the magnetic field within the solenoid, given a constant current. This is because each turn of wire contributes to the total magnetic field, and more turns mean more superposition of the magnetic fields from each individual turn, resulting in a stronger aggregate field.

In our problem, calculating the value of \(n\) was instrumental for figuring out the total number of turns in the solenoid. First, we computed \(n\) and then multiplied it by the solenoid's length to obtain the total number of turns. This relationship showcases how relevant and interconnected the physical dimensions of a solenoid and its electromagnetic properties are.

Converting Units in Physics

Converting units is a fundamental skill in solving physics problems, ensuring consistency and accuracy in calculations. For the solenoid exercise, we had to convert millimeters to meters and milliteslas to teslas in order to use consistent SI units (International System of Units) in our formula. This step is crucial; without it, the formula would yield an incorrect value for the magnetic field.

Always pay close attention to units when performing physics calculations and convert them to a common system before applying any formulas. This keeps computations error-free and ensures the calculated quantities' physical significance. A misstep in unit conversion can lead to large-scale errors, which makes it one of the most important procedural steps in physics problem-solving.

The simple act of converting \(5.0 \, \mathrm{mm}\) to \(5.0 \times 10^{-3} \, \mathrm{m}\) and \(5.0 \, \mathrm{mT}\) to \(5.0 \times 10^{-3} \, \mathrm{T}\) in the given problem is a testament to the importance of ensuring that all quantities are expressed in compatible units before proceeding with calculations.