We have seen that a long solenoid produces a uniform magnetic field directed along the axis of a cylindrical region. To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region, however, one can use the saddle coils illustrated in Figure P29.46. The loops are wrapped over a long, somewhat flattened tube. Figure P29.46a shows one wrapping of wire around the tube. This wrapping is continued in this manner until the visible side has many long sections of wire carrying current to the left in Figure P29.46a and the back side has many lengths carrying current to the right. The end view of the tube in Figure P29.46b shows these wires and the currents they carry. By wrapping the wires carefully, the distribution of wires can take the shape suggested in the end view such that the overall current distribution is approximately the superposition of two overlapping, circular cylinders of radius R (shown by the dashed lines) with uniformly distributed current, one toward you and one away from you. The current density J is the same for each cylinder. The center of one cylinder is described by a position vector d → relative to the center of the other cylinder. Prove that the magnetic field inside the hollow tube is μ 0 Jd /2 downward. Suggestion: The use of vector methods simplifies the calculation.
We have seen that a long solenoid produces a uniform magnetic field directed along the axis of a cylindrical region. To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region, however, one can use the saddle coils illustrated in Figure P29.46. The loops are wrapped over a long, somewhat flattened tube. Figure P29.46a shows one wrapping of wire around the tube. This wrapping is continued in this manner until the visible side has many long sections of wire carrying current to the left in Figure P29.46a and the back side has many lengths carrying current to the right. The end view of the tube in Figure P29.46b shows these wires and the currents they carry. By wrapping the wires carefully, the distribution of wires can take the shape suggested in the end view such that the overall current distribution is approximately the superposition of two overlapping, circular cylinders of radius R (shown by the dashed lines) with uniformly distributed current, one toward you and one away from you. The current density J is the same for each cylinder. The center of one cylinder is described by a position vector d → relative to the center of the other cylinder. Prove that the magnetic field inside the hollow tube is μ 0 Jd /2 downward. Suggestion: The use of vector methods simplifies the calculation.
Solution Summary: The author explains the magnetic field inside the hollow tube and the position vector of center of one cylinder relative to another.
We have seen that a long solenoid produces a uniform magnetic field directed along the axis of a cylindrical region. To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region, however, one can use the saddle coils illustrated in Figure P29.46. The loops are wrapped over a long, somewhat flattened tube. Figure P29.46a shows one wrapping of wire around the tube. This wrapping is continued in this manner until the visible side has many long sections of wire carrying current to the left in Figure P29.46a and the back side has many lengths carrying current to the right. The end view of the tube in Figure P29.46b shows these wires and the currents they carry. By wrapping the wires carefully, the distribution of wires can take the shape suggested in the end view such that the overall current distribution is approximately the superposition of two overlapping, circular cylinders of radius R (shown by the dashed lines) with uniformly distributed current, one toward you and one away from you. The current density J is the same for each cylinder. The center of one cylinder is described by a position vector
d
→
relative to the center of the other cylinder. Prove that the magnetic field inside the hollow tube is μ0Jd/2 downward. Suggestion: The use of vector methods simplifies the calculation.
How would partial obstruction of an air intake port of an air-entrainment mask effect FiO2 and flow?
14
Z
In figure, a closed surface with q=b=
0.4m/
C =
0.6m
if the left edge
of the closed surface at position X=a,
if E is non-uniform and is given by
€ = (3 + 2x²) ŷ N/C, calculate the
(3+2x²)
net electric flux leaving the closed
surface.
Chapter 30 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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What is Electromagnetic Induction? | Faraday's Laws and Lenz Law | iKen | iKen Edu | iKen App; Author: Iken Edu;https://www.youtube.com/watch?v=3HyORmBip-w;License: Standard YouTube License, CC-BY