A closed curve encircles several conductors. The line integral ∮ B → ⋅ dl → around this curve is 3.83 × 10 −4 T • m. (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.
A closed curve encircles several conductors. The line integral ∮ B → ⋅ dl → around this curve is 3.83 × 10 −4 T • m. (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.
A closed curve encircles several conductors. The line integral
∮
B
→
⋅
dl
→
around this curve is 3.83 × 10−4 T • m. (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.
Four long, parallel conductors carry equal currents of I = 4.00 A. The figure below is an end view of the conductors. The current direction is into the page at points A and B and out of the page at C and D.
Already calculated B= 16 uT . But there is another question I dont know how to answer:
What If? What would be the magnitude and direction of the initial acceleration of an electron moving with velocity 3.18 ✕ 105 m/s into the page at point P?ays:
(magnitude and direction)
The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
A closed curve encircles several conductors. The line integral PB.dL around this curve
is 3,83x104 T.m
(a) What is the net current in the conductors?
(b) If you were to integrate around the curve in the opposite direction, what would be
the value of the line integral?
Select one:
lenci=D
305 A, 0.0 T.m
lencl = 502 A, -3.83 x 10 4T.m
lenci = 502 A, -7.66 x 104 T.m
lencl = 600 A, -7.66 x 104T.m
lencl = 305 A, -3.83 x 10 4 T.m
lencl =
502 A, 0.0 T.m
%3D
Chapter 28 Solutions
University Physics with Modern Physics, Volume 2 (Chs. 21-37); Mastering Physics with Pearson eText -- ValuePack Access Card (14th Edition)
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