Skip to main content
Ch 29: The Magnetic Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 29, Problema 82b

A long, straight conducting wire of radius R has a nonuniform current density J = J₀r/R, where J₀ is a constant. The wire carries total current I. Find an expression for the magnetic field strength inside the wire at radius r.

Guida verificata passo dopo passo
1
Start by recalling Ampère's Law, which states: ∮B·dl = μ₀I_enc, where B is the magnetic field, dl is the infinitesimal length element of the loop, μ₀ is the permeability of free space, and I_enc is the current enclosed by the Amperian loop.
Since the magnetic field inside the wire is symmetric and circular around the axis of the wire, choose a circular Amperian loop of radius r (where r < R) centered on the wire's axis. The magnetic field B will have the same magnitude at all points on this loop, and the path length of the loop is its circumference, 2πr.
To find the enclosed current I_enc, integrate the current density J over the cross-sectional area enclosed by the Amperian loop. The current density is given as J = J₀r/R. The infinitesimal current element is dI = J dA, where dA = 2πr' dr' (the area of a thin ring at radius r').
Set up the integral for I_enc: I_enc = ∫(J₀r'/R) (2πr' dr') from 0 to r. Simplify the integral to I_enc = (2πJ₀/R) ∫(r'^2 dr') from 0 to r. Evaluate the integral to find I_enc = (2πJ₀/R) (r³/3).
Substitute I_enc into Ampère's Law: B(2πr) = μ₀(2πJ₀/R)(r³/3). Simplify to find the magnetic field strength inside the wire: B = (μ₀J₀r²)/(3R).

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
7m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Current Density

Current density (J) is defined as the electric current (I) flowing per unit area (A) of a cross-section of a conductor. In this case, the current density is nonuniform and varies with the radial position (r) within the wire, given by the equation J = J₀r/R. Understanding current density is crucial for calculating the total current and the resulting magnetic field.
Video consigliato:
Percorso guidato
8:13
Intro to Density

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 expressed mathematically as ∮B·dl = μ₀I_enc, where B is the magnetic field, dl is a differential length element, μ₀ is the permeability of free space, and I_enc is the enclosed current. This law is essential for deriving the magnetic field inside the wire based on the current density.
Video consigliato:

Magnetic Field Inside a Conductor

The magnetic field inside a conductor carrying current can be determined by considering the distribution of current density and applying Ampère's Law. For a cylindrical conductor, the magnetic field strength (B) at a distance r from the center can be derived by integrating the contributions of the current density over the area enclosed by the radius r. This concept is key to solving the problem of finding the magnetic field strength at a specific radius within the wire.
Video consigliato:
Percorso guidato
07:43
Electric Fields in Conductors