**Problem Statement:**Two long parallel wires, 79.4 cm apart, are carrying currents of 12.6 A and 28.1 A in the opposite direction. What is the magnitude of the magnetic field halfway between the wires?**Explanation:**To solve this problem, we use the formula for the magnetic field due to a long straight current-carrying wire:\[ B = \frac{\mu_0 \cdot I}{2 \pi \cdot d} \]Where:- \( B \) is the magnetic field,- \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)),- \( I \) is the current in the wire,- \( d \) is the distance from the wire to the point where the field is being calculated.Since the currents are in opposite directions, their magnetic fields will be in opposite directions at the midpoint. Calculate the magnetic field due to each wire at the midpoint, then add them vectorially, considering their directions.1. **Calculate the distance from each wire to the midpoint:** \[ d = \frac{79.4 \, \text{cm}}{2} = 39.7 \, \text{cm} = 0.397 \, \text{m} \]2. **Calculate the magnetic field due to each wire:** \[ B_1 = \frac{4\pi \times 10^{-7} \cdot 12.6}{2\pi \cdot 0.397} \] \[ B_2 = \frac{4\pi \times 10^{-7} \cdot 28.1}{2\pi \cdot 0.397} \]3. **Determine the net magnetic field:** Since the fields are in opposite directions: \[ B_{\text{net}} = |B_2 - B_1| \]Compute these values to find the magnitude of the net magnetic field halfway between the wires.

Given data: Distance between the wire (d) = 79.4 cm = 0.794 m Currents in the wires I1 = 12.6 A and I2 = 28.1 A The currents are in opposite direction Required: The magnitude of the magnetic field midway between the wiresWe know that Permeability of free space μ0 = 4π×10-7 H/m The magnitude of the magnetic field due to a long current carrying wire at distance r is given by B = μ0I2πr The distance midpoint from the wires r = d2= 0.794 m2= 0.397 m The currents are opposite direction so the direction of the magnetic force halfway between the wires are in same direction. The magnitude of the magnetic field halfway between the wires B = B1 + B2= μ0I12πr + μ0I22πr= μ02πr(I1 + I2) = 4π×10-7 H/m2π×(0.397 m)(12.6 A + 28.1 A)= 2.05×10-5 T The magnitude of the magnetic field halfway between the wires is 2.05×10-5 T

Answered: Two long parallel wires 79.4 cm apart…

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Two long parallel wires 79.4 cm apart are carrying currents of 12.6 A and 28.1 A in the opposite direction. What is the magnitude of the magnetic field halfway between the wires?

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter24: Magnetic Fields

Section: Chapter Questions

Problem 103A

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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

**Problem Statement:**

Two long parallel wires, 79.4 cm apart, are carrying currents of 12.6 A and 28.1 A in the opposite direction. What is the magnitude of the magnetic field halfway between the wires?

**Explanation:**

To solve this problem, we use the formula for the magnetic field due to a long straight current-carrying wire:

\[ B = \frac{\mu_0 \cdot I}{2 \pi \cdot d} \]

Where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)),
- \( I \) is the current in the wire,
- \( d \) is the distance from the wire to the point where the field is being calculated.

Since the currents are in opposite directions, their magnetic fields will be in opposite directions at the midpoint. Calculate the magnetic field due to each wire at the midpoint, then add them vectorially, considering their directions.

1. **Calculate the distance from each wire to the midpoint:**
\[ d = \frac{79.4 \, \text{cm}}{2} = 39.7 \, \text{cm} = 0.397 \, \text{m} \]

2. **Calculate the magnetic field due to each wire:**
\[ B_1 = \frac{4\pi \times 10^{-7} \cdot 12.6}{2\pi \cdot 0.397} \]
\[ B_2 = \frac{4\pi \times 10^{-7} \cdot 28.1}{2\pi \cdot 0.397} \]

3. **Determine the net magnetic field:**
Since the fields are in opposite directions:
\[ B_{\text{net}} = |B_2 - B_1| \]

Compute these values to find the magnitude of the net magnetic field halfway between the wires.

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