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Ch 28: Sources of Magnetic Field
Chapter 28, Problem 28

. 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. (a) 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?

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1
Identify the direction of the currents in both wires. If the currents are in the same direction, the force between the wires will be attractive. If the currents are in opposite directions, the force will be repulsive.
Use Ampere's Law to determine the magnetic field produced by one wire at the location of the other. The magnetic field, B, due to a long straight wire carrying a current I at a distance r is given by the formula B = \(\frac{\mu_0 I}{2\pi r}\), where \(\mu_0\) is the permeability of free space.
Calculate the force on a segment of the second wire due to the magnetic field produced by the first wire. The force per unit length, f, on a wire carrying a current I in a magnetic field B is given by f = I B. Use the length of the wire segment given in the problem to find the total force.
Substitute the values for the currents, the distance between the wires, and the length of the wire segment into the formulas to find the magnetic field and the force.
Determine the nature of the force (attractive or repulsive) based on the direction of the currents and summarize the findings.

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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 current-carrying wire generates a magnetic field around it, which can be described by the right-hand rule. The strength of this magnetic field 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. This magnetic field interacts with other currents, leading to forces between parallel wires.
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Force Between Parallel Current-Carrying Wires

When two parallel wires carry currents, they exert forces on each other due to their magnetic fields. The force per unit length between the wires can be calculated using the formula F/L = (μ₀I₁I₂)/(2πd), where F is the force, L is the length of the wire, I₁ and I₂ are the currents, and d is the distance between the wires. The direction of the force is determined by the direction of the currents: if they flow in the same direction, the force is attractive; if in opposite directions, it is repulsive.
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Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic field and the force on a current-carrying conductor. For a straight wire, if you point your thumb in the direction of the current, your fingers curl in the direction of the magnetic field lines. This rule helps visualize the interactions between currents and magnetic fields, essential for predicting the behavior of the wires in the problem.
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Related Practice
Textbook Question
A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A. (b) What total length of wire is required?
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Textbook Question
Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is 0.50 G. Is this value cause for worry?
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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.
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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. (b) 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?
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Textbook Question
A solenoid 25.0 cm long and with a cross-sectional area of 0.500 cm^2 contains 400 turns of wire and carries a current of 80.0 A. Calculate: (c) the total energy contained in the coil's magnetic field (assume the field is uniform);
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