Problem 11 - University Physics with Modern Physics

Question

A long, straight wire lies along the \(z\)-axis and carries a 4.00-A current in the \(+z\)-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire centered at the origin: (a) \(x =\) 2.00 m, \(y = 0\), \(z =\) 0; (b) x = 0, \(y =\) 2.00 m, \(z =\) 0; (c) \(x =\) 2.00 m, \(y =\) 2.00 m, \(z =\) 0; (d) \(x\) = 0, \(y\) = 0, \(z\) = 2.00 m,

Step-by-Step Solution

Answer

The magnetic fields are perpendicular to the line joining the wire and the observation points, with directions determined via the right-hand rule. Magnitude for (a) is approximately 0.5 nT perpendicular and various direction results follow for (b), (c) encouraging cross elements of position vectors use.

Step 1: Identify the Problem at Hand

We need to determine the magnetic field created by a thin wire at various positions. The wire is aligned with the z-axis and carries a current of 4.00 A in the positive z-direction. The magnetic field will be determined using the Biot-Savart Law, which is most appropriate for estimating the field near a finite segment of current.

Step 2: Apply the Biot-Savart Law

The Biot-Savart Law gives the magnetic field due to a current element as: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \cdot d\mathbf{s} \times \mathbf{r}}{r^3} \]where \(d\mathbf{B}\) is the magnetic field generated by a current element \(Id\mathbf{s}\), \(\mathbf{r}\) is the position vector from the current element to the observation point, and \(\mu_0\) is the permeability of free space.

Step 3: Calculate for Point (a)

For point (a) \((x = 2.00\, \text{m}, y = 0, z = 0)\), - The segment \(d\mathbf{s}\) is along \(z\)-axis: \(d\mathbf{s} = dz\hat{k}\).- The position vector is \(\mathbf{r} = (2\hat{i})\).- \(d\mathbf{s} \times \mathbf{r} = dz(-\hat{j})\), gives the direction of the magnetic field.- Calculate the magnitude using: \[ \|d\mathbf{B} \| = \frac{\mu_0 I dz}{4 \pi} \frac{2}{2^3} = \frac{\mu_0 I dz}{16 \pi} \]- Substituting values and constants, use \(\mu_0 = 4\pi \times 10^{-7}\, \text{T}\cdot\text{m/A}\), \(I = 4.00\, \text{A}\), and \(dz = 0.5\, \text{mm} = 0.5\times 10^{-3}\, \text{m}\), resulting in: \[ dB = \frac{4\times 10^{-7} \cdot 4 \cdot 0.5\times 10^{-3}}{16 \pi} = 0.5\times 10^{-10}\, \text{T}\].

Step 4: Direction for Point (a)

The direction of \(d\mathbf{B}\) according to the right-hand rule from \(d\mathbf{s} \times \mathbf{r}\) is in the negative y-direction (\(-\hat{j}\)).

Step 5: Calculate for Other Points

Repeat similar vector algebra for points (b), (c), and (d):- **Point (b)**: At \((x = 0, y = 2.00\, \text{m}, z = 0)\); \( \mathbf{r} = (0\hat{i} + 2\hat{j})\) gives a direction of \(+\hat{i}\).- **Point (c)**: At \((x = 2.00\, \text{m}, y = 2.00\, \text{m}, z = 0)\); \( \mathbf{r} = (2\hat{i} + 2\hat{j})\) results in resultant co-planar but different by symmetry.- **Point (d)**: At \((x = 0, y = 0, z = 2.00\, \text{m})\), \( \mathbf{r} = (2\hat{k})\), results in net zero field as magnetic fields created by symmetric sections cancel out.

Key Concepts

Magnetic Field Calculation

When calculating the magnetic field produced by a current-carrying wire, we often rely on a fundamental principle known as the Biot-Savart Law. This law helps us compute the magnetic field at different positions around the wire due to a small segment of current.

Given by the formula: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \cdot d\mathbf{s} \times \mathbf{r}}{r^3} \]where:

• \(d\mathbf{B}\) represents the magnetic field due to the small current element \(Id\mathbf{s}\).
• \(\mu_0\) is the permeability of free space, a constant value approximately \(4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}\).
• \(\mathbf{r}\) is the position vector pointing from the current segment to the point where the field is calculated.

The magnitude of the magnetic field depends on both the strength of the current and the geometry of the current element relative to the point in space you're evaluating.

Biot-Savart Law is essential for problems where you need precise field values close to small segments of current, and it guides through calculating fields that aren't symmetrical or uniformly distributed.

Current-Carrying Wire

A current-carrying wire, especially a straight one, forms the basis of many magnetic field calculations in physics. This is because the movement of electric charge generates a magnetic field perpendicular to the direction of the current flow.

For example, in the given exercise, we have a wire aligned along the z-axis carrying a 4.00 A current. This current flow creates a circular magnetic field around it, detectable in planes perpendicular to the wire. Each segment of the wire contributes to the magnetic field at various points in space.

When dealing with such problems:

• Consider the orientation of the wire and the current direction, as these influence the direction of the magnetic field.
• Utilize the right-hand rule to determine the field direction; when your thumb points in the direction of current, your curled fingers show the direction of the magnetic field.
• Keep in mind that symmetry in the setup can simplify calculations, as seen in some sub-parts of the exercise where components cancel each other resulting in zero net field.

This understanding is crucial for tackling more complex scenarios involving multiple wires or non-straight configurations.

Vector Algebra

Vector algebra plays a critical role in calculating magnetic fields, as seen with the Biot-Savart Law and dealing with forces, fields, and motions in three-dimensional space. The mathematics of vectors, including dot products and cross products, are foundational tools.

In the exercise, each point required considering the position vector \(\mathbf{r}\), which determines the direction from the wire segment to the point of interest.

• The cross product \(d\mathbf{s} \times \mathbf{r}\) provides the direction of the magnetic field.
• Magnitude of vectors and their distances (\(r^3\) in the denominator) affect the computed field strength as well.

Understanding vector algebra involves:

• Break down complex problems into smaller, manageable components for calculations.
• Use vector relationships and properties to deduce the direction of forces or fields, essential for interpreting results physically.

Consequently, mastering these vector principles is essential for engaging with physics problems that span forces and fields in spaces with multiple dimensions.