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29. Sources of Magnetic Field

Magnetic Force Between Parallel Currents

Physics

29. Sources of Magnetic Field

Magnetic Force Between Parallel Currents

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5:28 minutes

Problem 19

Textbook Question

(II) Let two long parallel wires, a distance d apart, carry equal dc currents I in the same direction. One wire is at 𝓍 = 0, the other at 𝓍 = d, Fig. 28–41. Determine $\vec{B}$ along the 𝓍 axis between the wires as a function of 𝓍.

Verified step by step guidance

Step 1: Understand the problem. Two long parallel wires separated by a distance d carry equal currents I in the same direction. We are tasked with finding the magnetic field B⃗ along the x-axis between the wires as a function of x, where x is the position along the axis between the wires.

Step 2: Recall the formula for the magnetic field produced by a long straight current-carrying wire at a distance r from the wire. The magnetic field is given by:

B = \frac{μ ₀ I}{2 π r}

, where μ₀ is the permeability of free space, I is the current, and r is the distance from the wire.

Step 3: Consider the contributions to the magnetic field at a point x along the x-axis from both wires. The wire at x = 0 produces a magnetic field at x due to the current I, and the wire at x = d also produces a magnetic field at x. Use the right-hand rule to determine the direction of the magnetic fields from each wire.

Step 4: Write the expressions for the magnetic fields from each wire. For the wire at x = 0, the distance to the point x is |x|, so the magnetic field is:

B

₁ = μ₀I2πx. For the wire at x = d, the distance to the point x is |d - x|, so the magnetic field is:

B

₂ = μ₀I2π|d-x|.

Step 5: Combine the magnetic field contributions. Since the currents are in the same direction, the magnetic fields from the two wires will oppose each other along the x-axis. The net magnetic field at a point x is the difference between the magnitudes of the two fields:

B = | B

₁ - B₂|. Substitute the expressions for B₁ and B₂ to find the net magnetic field as a function of x.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Field Due to a Current-Carrying Wire

A long straight wire carrying a current generates a magnetic field around it, described by Ampère's Law. The magnetic field strength (B) at a distance (r) from the wire is given by the formula B = (μ₀I)/(2πr), where μ₀ is the permeability of free space and I is the current. The direction of the magnetic field can be determined using the right-hand rule, which states that if you point your thumb in the direction of the current, your fingers curl in the direction of the magnetic field.

Superposition of Magnetic Fields

When multiple sources generate magnetic fields, the total magnetic field at a point is the vector sum of the individual fields. In this scenario, the magnetic fields produced by each wire will interact, and their contributions must be added together to find the resultant magnetic field at any point along the x-axis. This principle is crucial for determining the net magnetic field between the two wires, as the fields will have both magnitude and direction.

Distance and Magnetic Field Variation

The magnetic field strength varies with distance from the current-carrying wires. As you move along the x-axis between the two wires, the distance from each wire changes, affecting the magnetic field contributions from each. Specifically, the magnetic field will be stronger closer to the wire and weaker as you move away, which must be accounted for when calculating the total magnetic field at any given point between the wires.

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Related Practice

Textbook Question

Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I₁ and I₂ have the directions shown. Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other. Is the force attractive or repulsive?

Textbook Question

Four long, parallel power lines each carry 100 A currents. A cross-sectional diagram of these lines is a square, 20.0 cm on each side. For each of the three cases shown in Fig. E28.25, calculate the magnetic field at the center of the square.

A 175-g model airplane charged to 18.0 mC and traveling at 2.8 m/s passes within 7.8 cm of a wire, nearly parallel to its path, carrying a 25-A dc current. What acceleration (in g’s) does this interaction give the airplane?.

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Textbook Question

In Fig. 28–57 the top wire is 1.00-mm-diameter copper wire and is suspended in air due to the two magnetic forces from the bottom two wires. The current is 35.0 A in each of the two bottom wires. Calculate the required current in the suspended wire (M).

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Multiple Choice

Two very long wires of unknown lengths are a parallel distance of 2 m from each other. If both wires have 3 A of current flowing through them in the same direction, what must the force per unit length on each wire be?

BONUS:Is the mutual force between the wires attractive or repulsive?

Two long parallel wires lie in the x-y plane, and each carry

5.0 A

currents in opposite directions. Wire 1 lies along the

y = 3.0 cm

line and carries a current in the positive x direction; wire 2 lies along the y = 0

y = 0

line and carries a current in the negative x direction. What is the magnitude of the magnetic field at the point (

0 cm

1.5 cm

0 cm

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Textbook Question

Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I₁ and I₂ have the directions shown. Each current is doubled, so that I₁ becomes 10.0 A and I₂ becomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20 m length of the other?