Question

A long solenoid has a radius of \(3 \mathrm{~cm}, 5000\) turns \(/ \mathrm{m}\), and carries current \(I=0.25 \mathrm{~A} .\) The region \(0<\rho

Short answer

Question: Calculate the radius 'a' in a solenoid, given two different conditions on the total magnetic flux: (1) the total flux is 10μWb, (2) the flux is equally divided between two regions with different magnetic permeability. Answer: In both cases, the value of the radius 'a' is different. 1. For the total flux of 10μWb, the radius 'a' is approximately 1.4 x 10^-2 m. 2. For the equally divided magnetic flux between the two regions, the radius 'a' is approximately 2.2 x 10^-2 m.

Step-by-step solution

Write down the expression for the magnetic field in a solenoid

Since the magnetic field in the solenoid is parallel to the solenoid axis, we can consider only the axial component \(B_z\). The radial component \(B_\rho\) is irrelevant for the calculation of the magnetic flux. Using Ampere's Law, we find the axial magnetic field inside the solenoid (Region 1) given by: \[B_{z1} = \mu_{r1} \mu_0 n I\] Similarly, for Region 2, we have: \[B_{z2} = \mu_{r2} \mu_0 n I\]

Write down the expression for the magnetic flux in each region

The magnetic flux \(\Phi\) is given by the product of the magnetic field and the area through which it passes. Let's consider a cylindrical surface of radius \(\rho\) and height \(h\). The flux in each region can be defined as: \[\Phi_1 = B_{z1} (\pi(\rho^2-a^2)h)\] \[\Phi_2 = B_{z2} (\pi(a^2)h)\]

Solve the problem for case(a)

Given that the total flux is \(10\mu\text{Wb}\), we have: \[\Phi_{total} = \Phi_1 + \Phi_2 = 10\times10^{-6}\text{Wb}\] Substitute the expressions for \(B_{z1}\) and \(B_{z2}\) that we got in Step 1: \[(\mu_{r1}\mu_0nI)(\pi(\rho^2-a^2)h)+(\mu_{r2}\mu_0nI)(\pi(a^2)h)=10\times 10^{-6}\text{Wb}\] Using the given values, \(\mu_{r1}=5\), \(\mu_{r2}=1\), \(\rho=3\times10^{-2}\text{m}\), \(n=5000\text{turns/m}\), and \(I=0.25\text{A}\), we can solve for the unknown variable (a): \[a^2(4\pi\times10^{-7}\times 5000\times0.25)=5\times10^{-6}\] \[a^2 = \frac{5\times10^{-6}}{(4\pi\times10^{-7}\times 5000\times0.25)}\] Solving for "a", we get: \[a \approx 1.4\times10^{-2} \text{m}\]

Solve the problem for case(b)

In this case, it's given that the flux is equally divided between the regions \(0 < \rho < a\) and \(a < \rho < 3\text{cm}\). Therefore, we have: \[\Phi_1 = \Phi_2\] \[(\mu_{r1}\mu_0nI)(\pi(\rho^2-a^2)h) = (\mu_{r2}\mu_0nI)(\pi(a^2)h)\] Using the same given values as in case(a), we get: \[5(\rho^2-a^2) = a^2\] Since \(\rho = 3\times10^{-2}\text{m}\), we can solve for "a": \[a^2 = \frac{5(\rho^2)}{6}\] \[a^2 = \frac{5(3\times10^{-2})^2}{6}\] Solving for "a", we get: \[a \approx 2.2\times10^{-2} \text{m}\] The value of "a" found under both cases are as follows: Case(a): \(a \approx 1.4\times10^{-2} \text{m}\) Case(b): \(a \approx 2.2\times10^{-2} \text{m}\)

Key concepts

Solenoid

A solenoid is a coil of wire that is designed to create a magnetic field when an electric current flows through it. In a typical setup, a solenoid is constructed from many closely spaced loops or turns of wire. Each loop of wire contributes to the overall magnetic field, resulting in a relatively strong magnetic field inside the coil.
A key feature of a solenoid is its ability to produce a uniform magnetic field within its core. This property is particularly useful in various applications, including electromagnets, magnetic resonance imaging (MRI) machines, and transformers.
In the exercise, a long solenoid with many turns per meter is considered. The solenoid is carrying a current of 0.25 amperes and this current generates a magnetic field within the solenoid. This computed magnetic field helps calculate the magnetic flux inside different regions within the solenoid.

Magnetic Field

The magnetic field in a solenoid is one of its most important characteristics. When a current passes through the solenoid, the magnetic field is primarily confined within its core and is directed along the axis of the solenoid.
The strength of the magnetic field inside the solenoid can be calculated using Ampere's Law, which relates the magnetic field along a closed path to the current that runs through the path. For a solenoid, the formula used is:

• \( B = \mu n I \)

Here, \( \mu \) is the permeability, \( n \) is the number of turns per unit length, and \( I \) is the current. The uniform magnetic field ensures that the solenoid can effectively induce a magnetic flux, which plays a crucial role in the exercise section.
Additionally, this magnetic field creates two distinct regions within the solenoid, as defined in the problem. These regions have different values of permeability and hence different magnetic properties.

Permeability

Permeability is a measure of how easy it is for a material to support the formation of a magnetic field within itself. In simpler terms, it tells us how a material responds to the presence of a magnetic field.
The symbol for permeability is \( \mu \), and it can be divided into two components:

• \( \mu_0 \): The permeability of free space, a known constant \( (4\pi \times 10^{-7} \; \text{H/m}) \)
• \( \mu_r \): The relative permeability, which varies depending on the material

In the problem at hand, we have two regions with different relative permeabilities, \( \mu_r = 5 \) for the inner region and \( \mu_r = 1 \) for the outer region within the solenoid. This difference in permeability affects the magnetic flux distribution and is key to solving the problem.
Materials with higher permeability allow for a stronger magnetic field for the same amount of current, making them more efficient for use in applications involving magnetic circuits.

Ampere's Law

Ampere's Law is a fundamental concept in electromagnetism that relates the integrated magnetic field around a closed loop to the electric current passing through the loop. It is often used to find the magnetic field in symmetric situations, such as inside a solenoid.
The mathematical expression of Ampere's Law is:

• \( \oint B \cdot dl = \mu_0 I_{ ext{enc}} \)

Here \( \oint B \cdot dl \) is the closed line integral of the magnetic field \( B \) along the path, and \( I_{\text{enc}} \) is the current enclosed by the path.
In a solenoid, Ampere's Law simplifies the calculation because the magnetic field is uniform inside and negligible outside, allowing us to focus on the core of the solenoid. This simplification leads to the formula for the magnetic field as mentioned earlier: \( B = \mu n I \). This equation is essential for understanding the magnetic properties and calculating magnetic flux in devices like solenoids.