Question

28.39 The figure shows a cross section across the diameter of a long, solid, cylindrical conductor. The radius of the cylinder is \(R=10.0 \mathrm{~cm} .\) A current of \(1.35 \mathrm{~A}\) is uniformly distributed through the conductor and is flowing out of the page. Calculate the direction and the magnitude of the magnetic field at positions \(r_{\mathrm{a}}=0.0 \mathrm{~cm}\) \(r_{\mathrm{b}}=4.00 \mathrm{~cm}, r_{\mathrm{c}}=10.00 \mathrm{~cm}\) and \(r_{d}=16.0 \mathrm{~cm} .\)

Short answer

Question: Calculate and compare the magnetic fields at positions \(r_\text{a}=0.0 \mathrm{~cm},~ r_\text{b}=4.00 \mathrm{~cm},~ r_\text{c}=10.00 \mathrm{~cm}\) and \(r_\text{d}=16.0 \mathrm{~cm}\) in relation to the center of a cylindrical conductor with uniform current flowing out of the page. Use the current \(I = 1.35 \mathrm{~A}\) and the radius of the cylinder \(R = 10.0 \mathrm{~cm}\). Answer: The magnetic field at different positions are as follows: - At \(r_\text{a}\), the magnetic field is \(B_\text{a} = 0.0 \mathrm{~T}\). - At \(r_\text{b}\), the magnetic field is \(B_\text{b} = \frac{1}{2}\mu_{0}\left(\frac{1.35}{\pi(10)^{2}}\right)(4.00) \approx 5.42 \times 10^{-5} \mathrm{~T}\). - At \(r_\text{c}\), the magnetic field is \(B_\text{c} = \frac{1}{2}\mu_{0}\left(\frac{1.35}{\pi(10)^{2}}\right)(10.00) \approx 1.36 \times 10^{-4} \mathrm{~T}\). - At \(r_\text{d}\), the magnetic field is \(B_\text{d} = \frac{\mu_0 (1.35)}{2\pi(16.0)} \approx 5.37 \times 10^{-5} \mathrm{~T}\). The magnetic field is zero at the center of the conductor and increases as we move towards the surface. It reaches its maximum value on the surface and then decreases as we move away from the conductor. The magnetic field direction at each position is tangent to the circular loop, following the right-hand rule.

Step-by-step solution

State Ampere's Law

Ampère's Law states that the closed line integral of the magnetic field (\(\vec{B}\)) around a closed loop is equal to the product of the permeability of free space (\(\mu_0\)) and the enclosed current (I): $$\oint \vec{B} \cdot d\vec{l}=\mu_{0} I_{\text{enc}}$$ In this case, we have a cylindrical conductor, so we will use a circular loop with radius r as our Amperian loop.

Define the geometry of the problem

The current in the conductor is uniformly distributed, so the current density (J) can be found by dividing the total current (\(I\)) by the cross-sectional area (\(\pi R^2\)) of the conductor: $$J=\frac{I}{\pi R^{2}}$$ We can relate the enclosed current (\(I_{\text{enc}}\)) to the current density (J) by integrating the current density over some area: $$I_{\text{enc}}=J A$$

Apply Ampère's Law for different distances

We will find the magnetic field at the following positions: \(r_\text{a}=0.0 \mathrm{~cm},~ r_\text{b}=4.00 \mathrm{~cm},~ r_\text{c}=10.00 \mathrm{~cm}\) and \(r_\text{d}=16.0 \mathrm{~cm}\). We will consider three regions: 1. Inside conductor (\(r R\)) The magnitude of the magnetic field will be constant at each distance; however, its direction will change depending on the position.

Magnetic field inside the conductor (\(r

For \(r_\text{a}=0.0 \mathrm{~cm}\) and \(r_\text{b}=4.00 \mathrm{~cm}\), we are inside the conductor, so the enclosed area is a disk with a radius \(r\). The enclosed current can be calculated as $$I_{\text{enc}} = J\pi r^2$$ Now, applying Ampère’s Law to find the magnetic field, $$B \oint dl = \mu_0 I_{\text{enc}}$$ Since the magnetic field is tangent to the Amperian loop. and \(dl\) is in the same direction as \(\vec{B}\), \(B \oint dl = B (2\pi r)\). Substitute the enclosed current into Ampère's Law: $$B (2\pi r) = \mu_{0}\left(J\pi r^2\right)$$ Solving for \(B\): $$B(r) = \frac{1}{2}\mu_{0}\left(\frac{I}{\pi R^{2}}\right)r$$ Now, calculate the magnetic field at \(r_\text{a}\) and \(r_\text{b}\) using the current \(I = 1.35 \mathrm{~A}\) and the radius of the cylinder \(R = 10.0 \mathrm{~cm}\).

Magnetic field on the surface of the conductor (\(r=R\))

For \(r_\text{c}=10.00 \mathrm{~cm}\), we are on the surface of the conductor. Using the equation for the magnetic field inside the conductor and setting \(r=R\), we can find the magnetic field at \(r_\text{c}\): $$B(R) = \frac{1}{2}\mu_{0}\left(\frac{I}{\pi R^{2}}\right)R$$

Magnetic field outside the conductor (\(r>R\))

For \(r_\text{d} = 16.0 \mathrm{~cm}\), we are outside the conductor. The enclosed current is equal to the total current I. Applying Ampère's Law: $$B (2\pi r) = \mu_{0} I$$ Solving for \(B\): $$B(r) = \frac{\mu_{0} I}{2\pi r}$$ Now, calculate the magnetic field at \(r_\text{d}\) using the given current value.

Calculate the magnetic field

Using the above equations and given values, calculate the magnetic field for each distance. The direction of the magnetic field at each position will be tangent to the circular loop, following the right-hand rule.

Compare the results

Compare the magnetic fields at different positions to see how it changes as one moves away from the center of the conductor.

Key concepts

Cylindrical Conductor

A cylindrical conductor is an essential concept when discussing electricity and magnetism because it provides a medium for current to flow through a circular cross-section. This scenario is particularly relevant in solving magnetic field problems using Ampère's Law. In a cylindrical conductor, the current flows uniformly across its cross-section, and understanding this geometry is crucial for applying the right calculations.
To analyze the situation, imagine a long solid cylinder with a given radius, such as 10 cm. The uniform distribution of current across this surface results in a current density that remains constant and can be calculated using the formula:\[J = \frac{I}{\pi R^2}\]where \(J\) is the current density, \(I\) is the total current flowing out of the page, and \(R\) is the radius of the conductor.

Studying a cylindrical conductor allows us to explore how current is distributed within, on the surface, and outside the conductor, which then affects the resulting magnetic field.

Magnetic Field

The magnetic field created by a current-carrying cylindrical conductor is an intriguing element of electromagnetism. The presence of current flowing within the conductor gives rise to a magnetic field, which can be analyzed using Ampère's Law. Let's look at how the magnetic field behaves considering different positions relative to the conductor:

• **Inside the conductor:** For points well within the cylindrical conductor (such as \(r < R\)), the magnetic field at a distance \(r\) from the center is determined by the current enclosed within the loop of radius \(r\). Mathematically, we express it as:\[B(r) = \frac{1}{2}\mu_0\left(\frac{I}{\pi R^2}\right)r\]This indicates the linear increase in the magnetic field strength as we move from the center towards the edge of the conductor.
• **On the surface:** At the surface of the conductor (\(r = R\)), we use the same formula and replace \(r\) with \(R\) to find the magnetic field.\[B(R) = \frac{1}{2}\mu_0\frac{I}{R}\]
• **Outside the conductor:** Beyond the surface (for \(r > R\)), all the current is enclosed, thus the magnetic field decreases with the distance from the conductor:\[B(r) = \frac{\mu_0 I}{2\pi r}\]

It's crucial to apply the right-hand rule to determine the field's direction: thumb points in the direction of current, and fingers curl in the direction of the magnetic field lines.

Current Density

Current density is a fundamental principle in the study of electromagnetism, particularly when discussing current distribution across a conductor. It represents the amount of electrical current flowing per unit area of a cross-section through which it passes.
In the case of a cylindrical conductor, the current density \(J\) is consistently spread across the entire cross-section due to its uniform distribution and can be calculated with:\[J = \frac{I}{\pi R^2}\]where \(J\) is the current density, \(I\) is the total current, and \(\pi R^2\) is the area of the cylinder's cross section.

Understanding current density is vital because it helps us determine how much current is present in a specific area within the conductor. Furthermore, knowing the current density allows us to calculate the magnetic field using Ampère's Law by integrating over a specific area to find the enclosed current \(I_{enc}\):\[I_{enc} = J \times A\]
where \(A\) represents the area through which the current flows. Thus, an accurate grasp of current density allows one to apply laws of electromagnetism effectively, such as calculating the magnetic field within and around the conductor based on how the current spreads.