Two long, parallel wires are separated by 7.17 cm and carry currents of 2.43 A and 4.69 A, respectively. Find the magnitude of the magnetic force F that acts on a 3.51 m length of either wire. F = N
Two long, parallel wires are separated by 7.17 cm and carry currents of 2.43 A and 4.69 A, respectively. Find the magnitude of the magnetic force F that acts on a 3.51 m length of either wire. F = N
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement:**
Two long, parallel wires are separated by 7.17 cm and carry currents of 2.43 A and 4.69 A, respectively. Find the magnitude of the magnetic force \( F \) that acts on a 3.51 m length of either wire.
\( F = \) [ ] N
---
**Explanation:**
This problem involves calculating the magnetic force between two parallel wires carrying current, using the formula for magnetic force per unit length between two current-carrying wires:
\[
F/L = \frac{\mu_0 \cdot I_1 \cdot I_2}{2\pi \cdot d}
\]
where:
- \( F \) is the force,
- \( L \) is the length of the wire,
- \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)),
- \( I_1 \) and \( I_2 \) are the currents in the wires,
- \( d \) is the distance between the wires.
The given length of the wire is 3.51 m.
This problem is an application of concepts from electromagnetism typically covered in physics courses related to electric current and magnetic field interactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1101034f-7e27-4909-aa23-dfb86d036d28%2F713c7b50-0579-4e7b-8230-e2f24f4a93f6%2Fztl32nj_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Two long, parallel wires are separated by 7.17 cm and carry currents of 2.43 A and 4.69 A, respectively. Find the magnitude of the magnetic force \( F \) that acts on a 3.51 m length of either wire.
\( F = \) [ ] N
---
**Explanation:**
This problem involves calculating the magnetic force between two parallel wires carrying current, using the formula for magnetic force per unit length between two current-carrying wires:
\[
F/L = \frac{\mu_0 \cdot I_1 \cdot I_2}{2\pi \cdot d}
\]
where:
- \( F \) is the force,
- \( L \) is the length of the wire,
- \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)),
- \( I_1 \) and \( I_2 \) are the currents in the wires,
- \( d \) is the distance between the wires.
The given length of the wire is 3.51 m.
This problem is an application of concepts from electromagnetism typically covered in physics courses related to electric current and magnetic field interactions.
Expert Solution
Step 1: Given information
Given the values:
I₁ = 2.43 A
I₂ = 4.69 A
L = 3.51m
d = 7.17cm = 0.0717 m
Where,
I₁ and I₂ are the currents in the two wires,
L is the length of the wire segment, and
d is the separation distance between the wires.
We need to find the magnitude of the magnetic force (F) that acts on a 3.51m length of either wire.
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