Question

A long, straight cylindrical tube of inner radius \(a\) and outer radius \(b\), carries a total current \(i\) uniformly across its cross section. Determine the magnitude of the magnetic field from the tube at the midpoint between the inner and outer radii.

Short answer

Answer: The magnitude of the magnetic field 'B' at the midpoint can be calculated using the following formula: \(B = \frac{\mu_0 i(r^2-a^2)}{2\pi r(b^2 - a^2)}\) where 'i' is the total current, 'a' and 'b' are the inner and outer radii, respectively, 'r' is the distance from the center of the cylindrical tube to the midpoint, and \(\mu_0\) is the permeability of free space.

Step-by-step solution

Determine the midpoint of the radii

Let's find the midpoint between the inner radius (a) and the outer radius (b). This can be calculated as: Midpoint = \(\frac{a+b}{2}\)

Find the current enclosed by the circular path at the midpoint

We need to find the current enclosed by the midpoint circular path. Since the total current 'i' is uniformly distributed across the cross-section, the current density 'J' can be defined as: \(J = \frac{i}{\pi(b^2 - a^2)}\) The cross-sectional area of the midpoint circular path can be calculated using the difference between the area of the outer circle and the area of the inner circle. Let's define 'r' as the distance from the center of the cylindrical tube to the midpoint: \(r=\frac{a+b}{2}\) The cross-sectional area at the midpoint is: \(A_{mid} = \pi r^2 - \pi a^2\) Now, to find the current enclosed by the circular path at the midpoint, multiply the current density 'J' by the area \(A_{mid}\): \(I_{enclosed} = J \times A_{mid}\) \(I_{enclosed} =\frac{i \times (\pi r^2 - \pi a^2)}{\pi(b^2 - a^2)}\)

Apply Ampere's Law

Now, we'll apply Ampere's Law which relates the magnetic field around a closed loop to the current enclosed by the loop: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}\) Since the magnetic field is tangential to the circular path and has a constant magnitude, we can rewrite Ampere's Law as: \(B \oint dl = \mu_0 I_{enclosed}\) \(L = 2\pi r\), where 'L' is the circumference of the circular path. So, we have: \(B \times (2\pi r) = \mu_0 I_{enclosed}\)

Solve for the magnetic field

Now, we'll solve for the magnetic field 'B' using the equation derived in step 3: \(B = \frac{\mu_0 I_{enclosed}}{2\pi r}\) \(B = \frac{\mu_0 \times \frac{i \times (\pi r^2 - \pi a^2)}{\pi(b^2 - a^2)}}{2\pi r}\) Simplify the equation and solve for 'B': \(B = \frac{\mu_0 i(r^2-a^2)}{2\pi r(b^2 - a^2)}\) This gives us the magnitude of the magnetic field 'B' at the midpoint between the inner and outer radii of the cylindrical tube.

Key concepts

Ampere's Law

Ampere's Law is a fundamental principle in electromagnetism that establishes a relationship between an electric current and the magnetic field it generates. This law can be described using the equation \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} \) where \( \vec{B} \) represents the magnetic field, \( d\vec{l} \) is an infinitesimal element of a closed loop, and \( \mu_0 \) is the permeability of free space. The term \( I_{enclosed} \) refers to the net current passing through the loop.

In practice, Ampere's Law allows us to calculate \( \vec{B} \) for highly symmetric situations, such as a straight wire or a solenoid. When applying Ampere's Law to solve problems, one often selects an Amperean loop—a hypothetical closed path—where the magnetic field is constant in magnitude and direction. This simplifies the integral into a product of the magnetic field magnitude and the length of the loop, as seen in our cylindrical tube problem.

Current Density

Current density, symbolized by \( J \) and measured in amperes per square meter (\(\frac{A}{m^2}\), is a vector quantity that represents the amount of electric current flowing per unit area of a cross-section. In more simple terms, it's how much current is spread over a given area of material.

For a uniform current distribution, current density is calculated by dividing the total current \( I \) by the area \( A \) through which the current is flowing: \( J = \frac{I}{A} \). This concept is vital in our textbook example to determine the current enclosed by the midpoint circular path within the cylindrical tube. When a problem mentions 'uniform current distribution,' it means that the current density is the same throughout the cross-sectional area, which allows us to rely on a simple proportional relationship to find currents in various regions.

Magnetic Field Calculation

Magnetic field calculation involves determining the strength and direction of the magnetic field generated by electric currents. We usually symbolize the magnetic field with the letter \( B \) and quantify it in units called teslas (T). The magnetic field at any point in space can be affected by the shape of the current-carrying conductor, and the properties of the space through which the magnetic field passes.

In the exercise, we calculate the magnetic field's magnitude at a specific point using Ampere's Law, which simplifies due to the symmetry of the problem. The calculated field is directly proportional to the enclosed current and inversely proportional to the distance from the center of the cylindrical tube. The simplified form of Ampere's Law, \( B \times (2\pi r) = \mu_0 I_{enclosed} \) is used to arrive at the final expression for \( B \) at the midpoint between the inner and outer radii of the tube.

Uniform Current Distribution

Uniform current distribution is an assumption commonly made in physics problems that simplifies calculations. It implies that current is spread evenly across the cross-sectional area of a conductor. This assumption allows us to use the total current and the total area to find the current density, and in turn, predict how the current and resulting magnetic field will behave under these conditions.

In our cylindrical tube exercise, the total current \( i \) is uniformly distributed over the cross-sectional area between the inner radius \( a \) and the outer radius \( b \) of the tube. Thus, the current density \( J \) is the same at every point in this area. With this information, we can determine the enclosed current at any given radius within the tube, which is crucial for applying Ampere's Law and calculating the magnetic field at the midpoint between \( a \) and \( b \). This uniform distribution provides a linear relationship that can be used to find out how much of the total current is within any sub-area of the cross-section.