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נבנה לולאת אמפר כמתואר באיור: חוק אמפר- in B dl 4 KI .הצלעות המאונכות לציר הסיבוב מאונכות גם לכיוון השדה ולכן לא יתרמו לסירקולציה (לאינטגרל) .הצלע הרחוקה לא תתרום גם היא (ניתן לקחת אותה לאינסוף ,שם השדה אפס) .מכאן ,שהתרומה היחידה לאינטגרל תהיה מהצלע הפנימית: B dl BL הזרם שחותך את מישור הלולאה יהיה: dq d dq RLd RL RL dt dt I in ואז לבסוף- Bz L 4 K RL Bz 4 K R Magnetic Field Submitted by: I.D. 036553790 The problem: There is a infinite wire with a radius a carrying current I. At the distance d from the center of the wire there is a cylinder cavity with a radius b. Find the magnetic field inside the cavity. The solution: First let’s define the constant in SI: K = µ0 /4π (1) Now let’s find the current density: I = J(πa2 − πb2 ) I J~ = ẑ 2 π(a − b2 ) (2) (3) Due to the symmetry the magnetic field is in the φ̂ direction. We use super position with a full wire minus the cavity so inside a full wire the magnetic field is (where r is the distance from the center of the cylinder) I Z ~ ~ B1 · d` = 4πK J~ · d~s (4) B1 · 2πr = 4πKJπr2 (5) B1 = 2πKJr (6) Therefore, in vector notations ~ 1 = 2πKJ~r × ẑ B (7) The magnetic field inside the cavity is (where r0 is the distance from the center of the cavity): I Z ~ ~ B2 · d` = 4πK J~ · d~s (8) B2 · 2πr0 = 4πKJπr02 B2 = 2πKJr (9) 0 (10) Since ~r0 = ~r − d~ ~ × ẑ ~ 2 = −2πKJ(~r − d) B (11) so the total magnetic field is: ~ × ẑ = 2πKJ d~ × ẑ ~ total = B ~1 + B ~ 2 = 2πKJ~r × ẑ − 2πKJ(~r − d) B (12) Finally ~ total = B µ0 I µ0 Id d~ × ẑ = φ̂ 2 π(a2 − b2 ) 2 π(a2 − b2 ) 1 (13)