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Ch. 28 - Sources of Magnetic Field
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 28 - Sources of Magnetic FieldProblema 34
Capitolo 27, Problema 34

(III) A coaxial cable consists of a solid inner conductor of radius R1, surrounded by a concentric cylindrical tube of inner radius R2 and outer radius R3 (Fig. 28–45). The conductors carry equal and opposite currents I₀ distributed uniformly across their cross sections. Determine the magnetic field at a distance R from the axis for: (a) R < R1; (b) R1 < R < R2; (c) R2 < R < R3; (d) R > R3. (e) Let I₀ = 1.50 A, R1 = 1.00 cm , R2 = 2.00 cm , and R3 = 2.50 cm Graph B from R = 0 to R = 3.00 cm.

Guida verificata passo dopo passo
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Step 1: Understand the problem and identify the regions where the magnetic field needs to be calculated. The coaxial cable consists of a solid inner conductor and a cylindrical outer conductor. The magnetic field depends on the distance R from the axis and the current distribution. Use Ampère's Law, which states: ∮𝐵·𝑑𝑙 = μ₀𝐼ₑ𝑛𝑐, where μ₀ is the permeability of free space and Iₑ𝑛𝑐 is the enclosed current.
Step 2: For region (a) R < R₁: In this region, the distance R is within the solid inner conductor. The enclosed current Iₑ𝑛𝑐 is proportional to the area of the circle of radius R. Since the current is uniformly distributed, calculate Iₑ𝑛𝑐 as Iₑ𝑛𝑐 = I₀ × (πR² / πR₁²). Substitute this into Ampère's Law to find the magnetic field B.
Step 3: For region (b) R₁ < R < R₂: In this region, the distance R is in the gap between the inner conductor and the inner surface of the cylindrical tube. The enclosed current Iₑ𝑛𝑐 is equal to the total current I₀ carried by the inner conductor. Use Ampère's Law to calculate the magnetic field B, noting that the current distribution in this region does not contribute additional current.
Step 4: For region (c) R₂ < R < R₃: In this region, the distance R is within the cylindrical tube. The outer conductor carries a current of -I₀ (opposite to the inner conductor). The enclosed current Iₑ𝑛𝑐 is calculated by considering the contribution of the inner conductor and the portion of the outer conductor up to radius R. Use Ampère's Law to find B.
Step 5: For region (d) R > R₃: In this region, the distance R is outside the coaxial cable. The total enclosed current is zero because the inner conductor carries I₀ and the outer conductor carries -I₀. Therefore, the magnetic field B is zero. For part (e), graph B as a function of R from R = 0 to R = 3.00 cm, using the results from parts (a) through (d).

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Ampère's Law

Ampère's Law relates the integrated magnetic field around a closed loop to the electric current passing through that loop. It is mathematically expressed as ∮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 enclosed current. This law is fundamental for analyzing magnetic fields in systems with symmetrical current distributions, such as coaxial cables.

Magnetic Field Inside a Conductor

The magnetic field inside a conductor carrying a uniform current is determined by the distribution of that current. For a solid conductor, the magnetic field increases linearly from the center to the surface, while outside the conductor, it behaves according to Ampère's Law. Understanding how the magnetic field varies with distance from the axis is crucial for solving the problem presented in the coaxial cable scenario.

Superposition Principle

The Superposition Principle states that in a linear system, the total response (magnetic field, in this case) at a given point is the sum of the responses caused by each individual source. In the context of the coaxial cable, this means that the magnetic fields produced by the inner and outer conductors can be calculated separately and then combined to find the total magnetic field at any point in space.
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Domanda del libro di testo

(II) Two long parallel wires 8.20 cm apart carry 19.5-A dc currents in the same direction. Determine the magnetic field vector at a point P, 12.0 cm from one wire and 13.0 cm from the other. See Fig. 28–43. [Hint: Use the law of cosines. See Appendix A or inside rear cover.]

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(II) An electron enters a uniform magnetic field B = 0.28 T at a 45° angle to B→\(\overrightarrow{B}\). Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 2.2 x 106 m/s. See Fig. 27–48.


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(II) A circular conducting ring of radius 𝑅 is connected to two exterior straight wires at two ends of a diameter (Fig. 28–47). The current I splits into unequal portions as shown (unequal resistance) while passing through the ring. What is B→\(\overrightarrow{B}\) at the center of the ring?


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(II) Two long wires are oriented so that they are perpendicular to each other. At their closest, they are 20.0 cm apart (Fig. 28–42). What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 18.0 A and the bottom one carries 12.0 A?

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(II) Consider a straight section of wire of length d, as in Fig. 28–51, which carries a current I. (a) Show that the magnetic field at a point P a distance 𝑅 from the wire along its perpendicular bisector is


B=μ0I2πRd(d2+4R2)12B = \(\frac{\mu_0 I}{2\pi R}\) \(\frac{d}{(d^2 + 4R^2)^{\frac{1}{2}\)}}


(b) Show that this is consistent with Example 28–10 for an infinite wire.

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(II) A wire is formed into the shape of two half circles connected by equal-length straight sections as shown in Fig. 28–48. A current I flows in the circuit clockwise as shown. Determine (a) the magnitude and direction of the magnetic field at the center, C, and (b) the magnetic dipole moment of the circuit.

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