所選文字: Figure shows a coaxial cable, widely used in electronics to minimize interference either with or from signals carried on the cable. The cable consists of an inner solid conductor of radius a and a hollow outer conductor of Inner radlus b and thickness c. The two conductors carry equal but opposite currents I, distributed uniformly over their cross-sectional areas. Find expressions for the magnetic field strength as a function of radial position r (a) within the inner conductor, (b) between the inner and outer conductors, (c) within the outer conductor, and (d) beyond the outer conductor. 解釋
Question
所選文字: Figure shows a coaxial cable, widely used in electronics to minimize interference either with or from signals carried on the cable. The cable consists of an inner solid conductor of radius a and a hollow outer conductor of Inner radlus b and thickness c. The two conductors carry equal but opposite currents I, distributed uniformly over their cross-sectional areas. Find expressions for the magnetic field strength as a function of radial position r (a) within the inner conductor, (b) between the inner and outer conductors, (c) within the outer conductor, and (d) beyond the outer conductor. 解釋
Solution
The magnetic field strength in a coaxial cable can be determined using Ampere's Law, which states that the integral of the magnetic field (B) around a closed loop is equal to the permeability of free space (μ0) times the current (I) enclosed by the loop.
(a) Within the inner conductor (r < a): The current enclosed by a loop of radius r is I_inner = I * (r^2 / a^2), because the current is uniformly distributed. Therefore, the magnetic field is B = μ0 * I_inner / (2πr) = μ0 * I * r / (2πa^2).
(b) Between the inner and outer conductors (a < r < b): The current enclosed by a loop of radius r is the total current in the inner conductor, I. Therefore, the magnetic field is B = μ0 * I / (2πr).
(c) Within the outer conductor (b < r < b+c): The current enclosed by a loop of radius r is I - I * ((r - b) / c) = I * (1 - (r - b) / c), because the current is uniformly distributed in the outer conductor. Therefore, the magnetic field is B = μ0 * I * (1 - (r - b) / c) / (2πr).
(d) Beyond the outer conductor (r > b+c): The current enclosed by a loop of radius r is zero, because the currents in the inner and outer conductors are equal and opposite. Therefore, the magnetic field is B = 0.
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