Problem 26 - Physics

Question

A \(45-\mathrm{m}\) length of wire is stretched horizontally between two vertical posts. The wire carries a current of \(75 \mathrm{~A}\) and experiences a magnetic force of \(0.15 \mathrm{~N}\). Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of \(60.0^{\circ}\) with respect to the wire.

Step-by-Step Solution

Answer

Earth's magnetic field magnitude is approximately \( 5.13 \times 10^{-5} \, \mathrm{T} \).

Step 1: Understand the Force-Current Relationship

The force exerted by a magnetic field on a current-carrying wire can be calculated using the formula: \[ F = I \cdot L \cdot B \cdot \sin(\theta) \]where \( F \) is the magnetic force, \( I \) is the current, \( L \) is the length of the wire, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the wire and the magnetic field.

Step 2: Substitute Known Values into the Formula

We know the magnetic force \( F = 0.15 \, \mathrm{N} \), the current \( I = 75 \, \mathrm{A} \), the length of the wire \( L = 45 \, \mathrm{m} \), and the angle \( \theta = 60.0^{\circ} \). Substitute these into the formula:\[ 0.15 = 75 \times 45 \times B \times \sin(60^{\circ}) \]

Step 3: Simplify the Expression

Calculate \( \sin(60^{\circ}) \), which is equal to \( \frac{\sqrt{3}}{2} \), and substitute:\[ 0.15 = 3375 \times B \times \frac{\sqrt{3}}{2} \]

Step 4: Solve for the Magnetic Field \( B \)

Rearrange the equation to solve for \( B \):\[ B = \frac{0.15}{3375 \times \frac{\sqrt{3}}{2}} \]Calculate the value:\[ B \approx \frac{0.15}{2922.46} \approx 5.13 \times 10^{-5} \, \mathrm{T} \]

Step 5: Conclusion

The magnitude of the earth's magnetic field at the location of the wire is approximately \( 5.13 \times 10^{-5} \, \mathrm{T} \).

Key Concepts

Current-Carrying Wire

Understanding the concept of a current-carrying wire is crucial, especially when dealing with magnetic fields. A current-carrying wire is essentially any wire through which electric current flows. In this particular exercise, the wire stretches horizontally between two posts and carries a current of 75 amperes (A).

When electrical current (which is a flow of electric charge) runs through the wire, it produces a magnetic field around it. This is a fundamental principle of electromagnetism known as Ampere's Law.

Key points about current-carrying wires:

• The direction of the current affects the direction of the magnetic force experienced.
• The magnitude of the magnetic force can change based on alterations in current, length of the wire, and its orientation in the magnetic field.
• The larger the current, the stronger the magnetic interaction.

Understanding these properties helps in analyzing and predicting the behavior of wires in magnetic fields.

Magnetic Field Strength

Magnetic field strength is a measure of how strong a magnetic field is at a given point. It's often represented by the symbol \( B \) and is measured in teslas (T).

In this problem, determining the magnetic field strength of the Earth requires understanding its relationship to the force experienced by a wire. The relevant formula for calculating the force on a wire due to a magnetic field is:
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]

This formula highlights a few important dependencies:

• Current \( I \)
• Length of the wire \( L \)
• Magnetic field strength \( B \)
• Angle of interaction \( \theta \)

In our exercise, we use known values and solve for \( B \) to find that Earth's magnetic field strength at the wire's location is approximately \( 5.13 \times 10^{-5} \mathrm{T} \). This shows how the magnetic field can be calculated using known quantities of force, current, wire length, and angle.

Angle of Interaction

The angle of interaction in our context is the angle between the magnetic field and the wire. In the formula \( F = I \cdot L \cdot B \cdot \sin(\theta) \), \( \theta \) stands for this specific angle.

This angle greatly influences the force experienced by the wire. Specifically, the magnetic force is maximized when the field is perpendicular to the current, meaning \( \theta = 90^{\circ} \). Conversely, when \( \theta = 0 \) or \( \theta = 180^{\circ} \), the force is nonexistent.

In our problem, the angle is \( 60^{\circ} \). Using this angle, we compute \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \). The term \( \sin(\theta) \) modifies the effective magnetic force experienced by the wire, thus making the angle a critical component of magnetic force calculations.

Physics Problem Solving

Approaching physics problems systematically can greatly simplify complex calculations. Let's discuss a simple step-by-step method to tackle these problems, using our exercise as a prime example.

1. **Understand the Problem:** Start by identifying known values and what you are asked to find. In this case, we aim to calculate Earth's magnetic field strength using given parameters.

2. **Identify the Relevant Formula:** Recognize the correct formula to apply. Here, \( F = I \cdot L \cdot B \cdot \sin(\theta) \) relates all given quantities to the unknown magnetic field strength.

3. **Substitute Known Values:** Plug in the values you know into the formula. This guides the calculation step by step.

4. **Solve for the Unknown:** Rearrange the equation as necessary to isolate the unknown factor, here the magnetic field strength \( B \).

5. **Evaluate the Conclusion:** Finally, after computing, confirm the unit and reasonableness of your solution. This reflection is crucial for ensuring the calculation aligns with physical expectations and units.

By adhering to these steps, we can systematically break down and solve physics problems effectively.