(I) A 2.5-mm-diameter copper wire carries a 28-A dc current (uniform across its cross section). Determine the magnetic field: (a) at the surface of the wire; (b) inside the wire, 0.50 mm below the surface; (c) outside the wire 2.5 mm from the surface.
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Ampere's Law (Calculus)
Problem 55
Textbook Question
Three long parallel wires are 3.5 cm from one another. (Looking along them, they are at three corners of an equilateral triangle.) The current in each wire is 9.50 A, but its direction in wire M is opposite to that in wires N and P (Fig. 28–57). Determine the magnetic force per unit length on each wire due to the other two.


1
Step 1: Understand the problem setup. The three wires form an equilateral triangle with a side length of 3.5 cm. The current in wire M flows in the opposite direction to the currents in wires N and P. The goal is to calculate the magnetic force per unit length on each wire due to the other two wires. The magnetic force between two parallel currents is given by the formula: \( F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \), where \( \mu_0 \) is the permeability of free space, \( I_1 \) and \( I_2 \) are the currents, and \( d \) is the distance between the wires.
Step 2: Calculate the magnetic force per unit length between any two wires. Since the triangle is equilateral, the distance \( d \) between any two wires is the same (3.5 cm = 0.035 m). Use the formula \( F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \) to find the magnitude of the force per unit length between two wires. Substitute \( \mu_0 = 4\pi \times 10^{-7} \ \text{T·m/A} \), \( I_1 = I_2 = 9.50 \ \text{A} \), and \( d = 0.035 \ \text{m} \).
Step 3: Determine the direction of the forces. Use the right-hand rule to determine the direction of the magnetic force between each pair of wires. For wire M, the forces due to wires N and P will have components that need to be resolved into horizontal and vertical directions. Similarly, for wires N and P, consider the forces due to the other two wires and their directions.
Step 4: Resolve the forces into components. For wire M, the forces due to wires N and P will form an angle of 60° with each other (since the triangle is equilateral). Resolve these forces into horizontal and vertical components using trigonometric functions: \( F_x = F \cos(60°) \) and \( F_y = F \sin(60°) \). Add the components vectorially to find the net force per unit length on wire M. Repeat this process for wires N and P.
Step 5: Combine the results. After resolving the forces into components and summing them, calculate the net magnetic force per unit length on each wire. Ensure that the directions of the forces are consistent with the right-hand rule and the geometry of the setup. The final result will give the magnitude and direction of the force per unit length on each wire.

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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 an electric current generates a magnetic field around it. The direction of the magnetic field can be determined using the right-hand rule, where the thumb points in the direction of the current and the curled fingers indicate the direction of the magnetic field lines. The strength of the 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.
Magnetic Force Between Parallel Wires
When two parallel wires carry currents, they exert magnetic forces on each other. The force per unit length between two parallel wires is given by the formula F/L = (μ₀I₁I₂)/(2πd), where I₁ and I₂ are the currents in the wires, d is the distance between them, and μ₀ is the permeability of free space. The direction of the force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.
Superposition Principle in Magnetism
The superposition principle states that the total magnetic force acting on a wire due to multiple other wires is the vector sum of the individual forces exerted by each wire. In this scenario, each wire experiences forces from the other two wires, and these forces must be calculated separately and then combined, taking into account their directions and magnitudes to find the net force per unit length on each wire.
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