8. A 10 cm length of wire carries a current of 5.0 A. The wire is in a uniform magnetic field with a strength of 10 mT, as in the figure. What is the magnitude and direction of the force on this segment of wire? X X XX B XX XXXX

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Chapter1: Units, Trigonometry. And Vectors
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**Problem 8:**

A 10 cm length of wire carries a current of 5.0 A. The wire is in a uniform magnetic field with a strength of 10 mT, as depicted in the figure. What is the magnitude and direction of the force on this segment of wire?

**Diagram Explanation:**

The diagram shows a vertical wire carrying current \( I \) upwards. The wire is positioned in a magnetic field \( \mathbf{B} \) represented by crosses ("X"), indicating the field is directed into the page.

### Solution Steps:

1. **Identify the formula for magnetic force**:
   The magnetic force on a current-carrying wire is given by:
   \[
   \mathbf{F} = I \cdot L \cdot \mathbf{B} \cdot \sin(\theta)
   \]
   Where \( I \) is the current, \( L \) is the length of the wire, \( \mathbf{B} \) is the magnetic field, and \( \theta \) is the angle between the wire and the magnetic field direction.

2. **Given values**:
   - Length of wire, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
   - Current, \( I = 5.0 \, \text{A} \)
   - Magnetic field strength, \( \mathbf{B} = 10 \, \text{mT} = 0.01 \, \text{T} \)

3. **Calculate the force**:
   Since the wire is perpendicular to the magnetic field (\( \theta = 90^\circ \)), \(\sin(\theta) = 1\).
   \[
   \mathbf{F} = 5.0 \, \text{A} \times 0.1 \, \text{m} \times 0.01 \, \text{T} \times 1 = 0.005 \, \text{N}
   \]

4. **Direction of the Force**:
   Using the right-hand rule, point your thumb in the direction of the current (upwards) and your fingers in the direction of the magnetic field (into the page). Your palm points in the direction of the force. Therefore, the force is directed to the left. 

In summary, the magnitude of
Transcribed Image Text:**Problem 8:** A 10 cm length of wire carries a current of 5.0 A. The wire is in a uniform magnetic field with a strength of 10 mT, as depicted in the figure. What is the magnitude and direction of the force on this segment of wire? **Diagram Explanation:** The diagram shows a vertical wire carrying current \( I \) upwards. The wire is positioned in a magnetic field \( \mathbf{B} \) represented by crosses ("X"), indicating the field is directed into the page. ### Solution Steps: 1. **Identify the formula for magnetic force**: The magnetic force on a current-carrying wire is given by: \[ \mathbf{F} = I \cdot L \cdot \mathbf{B} \cdot \sin(\theta) \] Where \( I \) is the current, \( L \) is the length of the wire, \( \mathbf{B} \) is the magnetic field, and \( \theta \) is the angle between the wire and the magnetic field direction. 2. **Given values**: - Length of wire, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \) - Current, \( I = 5.0 \, \text{A} \) - Magnetic field strength, \( \mathbf{B} = 10 \, \text{mT} = 0.01 \, \text{T} \) 3. **Calculate the force**: Since the wire is perpendicular to the magnetic field (\( \theta = 90^\circ \)), \(\sin(\theta) = 1\). \[ \mathbf{F} = 5.0 \, \text{A} \times 0.1 \, \text{m} \times 0.01 \, \text{T} \times 1 = 0.005 \, \text{N} \] 4. **Direction of the Force**: Using the right-hand rule, point your thumb in the direction of the current (upwards) and your fingers in the direction of the magnetic field (into the page). Your palm points in the direction of the force. Therefore, the force is directed to the left. In summary, the magnitude of
Expert Solution
Step 1

Given

current(I)=5 A

length(L)=10 cm=0.1 m

magnetic field (B)=10 mT=10-2 T

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