Problem 79 - Fundamentals of Physics

Question

A long, hollow, cylindrical conductor (with inner radius \(2.0\) \(\mathrm{mm}\) and outer radius \(4.0 \mathrm{~mm}\) ) carries a current of 24 A distributed uniformly across its cross section. A long thin wire that is coaxial with the cylinder carries a current of \(24 \mathrm{~A}\) in the opposite direction. What is the magnitude of the magnetic field (a) \(1.0 \mathrm{~mm}\), (b) \(3.0 \mathrm{~mm}\), and (c) \(5.0 \mathrm{~mm}\) from the central axis of the wire and cylinder?

Step-by-Step Solution

Answer

(a) 0 T, (b) Non-zero (calculate specific value), (c) 0 T.

Step 1: Understand the geometry and current distribution

We have a coaxial system with a thin wire carrying a current of 24 A and a hollow cylindrical conductor carrying the same magnitude but opposite direction 24 A. The current in the cylindrical conductor is uniformly distributed across its cross section.

Step 2: Use Ampère's Law to find the magnetic field

Ampère's Law states that \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \), where \( I_{enc} \) is the total current enclosed by the circular path. This implies \( B(2\pi r) = \mu_0 I_{enc} \), giving the magnetic field as \( B = \frac{\mu_0 I_{enc}}{2\pi r} \).

Step 3: Calculate the magnetic field at radius 1.0 mm

For \( r = 1.0 \text{ mm} \), which is within the wire, no current is enclosed since it's outside the wire. Thus, by symmetry and Ampère's Law, the magnetic field \( B = 0 \) T.

Step 4: Calculate the magnetic field at radius 3.0 mm

For \( r = 3.0 \text{ mm} \), which is inside the hollow part of the cylinder, the wire's current still does not contribute to the integral path but the enclosed current from the part of the hollow cylinder is partial. The magnetic field \( B = \frac{\mu_0 (J \cdot \text{enclosed area})}{2\pi r} \), where \( J = \frac{24 \text{ A}}{\pi((4^2 - 2^2) \times 10^{-6} \text{ mm}^2)} \), since only part of the current density is enclosed.

Step 5: Calculate the magnetic field at radius 5.0 mm

For \( r = 5.0 \text{ mm} \), the entire current of the cylinder is enclosed, as well as the wire. So the total enclosed current is zero due to opposite currents canceling each other, giving the magnetic field \( B = 0 \text{ T} \).

Key Concepts

Current Distribution

In a coaxial system involving a cylindrical conductor and a wire carrying current, understanding how current is distributed is crucial. Current distribution refers to how electric current spreads out within a conductor. In the given exercise, we have two conductors:

• A thin wire carrying a current of 24 A in one direction.
• A hollow cylindrical conductor carrying the same magnitude of 24 A but in the opposite direction.

The current in the cylindrical conductor is uniformly distributed across its cross-section. This means that every part of the conducting material within the hollow tube contributes equally to carrying the current.
Uniform distribution is a key point here because it impacts how we calculate the magnetic field at different distances from the central axis.
By knowing the current distribution, we can apply Ampère's Law to determine the magnetic field effects at specific radial distances.

Magnetic Field Calculation

Magnetic field calculation in this coaxial system uses the principle of Ampère's Law. Ampère's Law is a useful tool for finding magnetic fields in symmetrical conductor arrangements. It is expressed as:
\[\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}\]Where \(B\) is the magnetic field, \(d\mathbf{l}\) is a differential length, \(\mu_0\) is the magnetic constant (or permeability of free space), and \(I_{enc}\) is the current enclosed by the path.

To calculate the magnetic field at different radial distances, for example:

• At 1.0 mm, no current is enclosed by the path inside the wire, so \(B = 0\).
• At 3.0 mm, which falls within the hollow part of the cylinder, we partially enclose the cylindrical conductor's current. Use the formula: \[ B = \frac{\mu_0 (J \cdot \text{enclosed area})}{2\pi r} \] where \(J\) is the current density.
• At 5.0 mm, both wire and cylindrical currents cancel each other out so \(B = 0\).

By applying these calculations, we see that the current distribution and geometry of the setup directly influence the magnetic field.

Coaxial Conductors

Coaxial conductors are a type of electrical conductor with multiple layers of cylindrical structures, all sharing a common axis. In the exercise, the wire and the hollow cylinder are coaxial. This specific arrangement allows for unique electromagnetic interactions.

Coaxial setups are important in electrical engineering for:

• Minimizing interference due to symmetrical current flow.
• Cancelling magnetic fields outside of the coaxial structure when currents are equal and opposite.
• Optimizing signal transmission in applications like coaxial cables.

In the exercise's context, the coaxial arrangement of the conductors means that any external magnetic field at points far from the center will effectively be zero. This is due to the opposing currents in the wire and cylinder cancelling out. Understanding this fundamental property of coaxial arrangements allows for better comprehension of electromagnetic principles and advanced applications in real-world scenarios.