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Ch 28: Sources of Magnetic Field
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 15th EditionCh 28: Sources of Magnetic FieldProblema 39b
Capitolo 28, Problema 39b

A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c (Fig. E28.43). The central conductor and tube carry equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field at points outside the tube (r > c).

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Start by applying Ampère's Law, which relates the magnetic field around a closed loop to the current passing through the loop. The law is given by: ∮B·dl = μ₀I_enc, where B is the magnetic field, dl is a differential length element of the loop, μ₀ is the permeability of free space, and I_enc is the current enclosed by the loop.
Consider a circular Amperian loop of radius r, where r > c, centered on the axis of the conductors. Since the magnetic field is expected to be symmetric and tangential to the loop, the integral simplifies to B(2πr) = μ₀I_enc.
Determine the enclosed current I_enc. Since the currents in the central conductor and the tube are equal and opposite, the net enclosed current for r > c is zero. This is because the current in the central conductor is canceled by the current in the tube.
Substitute I_enc = 0 into the simplified Ampère's Law equation: B(2πr) = μ₀(0).
Solve for the magnetic field B. Since the right side of the equation is zero, the magnetic field B at points outside the tube (r > c) is zero. This result is consistent with the principle that the magnetic field outside a coaxial cable with equal and opposite currents is zero.

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Ampere's Law

Ampere's Law relates the integrated magnetic field around a closed loop to the electric current passing through the loop. It is expressed as ∮B·dl = μ₀I_enc, where B is the magnetic field, dl is a differential element of the loop, μ₀ is the permeability of free space, and I_enc is the enclosed current. This law is crucial for calculating magnetic fields in symmetric situations, such as the one described in the problem.
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Superposition Principle

The superposition principle states that the net magnetic field created by multiple sources is the vector sum of the magnetic fields produced by each source independently. In this problem, the magnetic fields generated by the central conductor and the conducting tube must be considered separately and then combined to find the total magnetic field at a point outside the tube.
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Magnetic Field of a Long Straight Conductor

The magnetic field around a long straight conductor carrying a current I is given by B = (μ₀I)/(2πr), where r is the radial distance from the conductor. This formula is derived from Ampere's Law and is essential for understanding how the magnetic field behaves around the central conductor and the conducting tube in the problem, especially when considering points outside the tube.
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