A solenoid 2.5 cm in diameter and 30 cm in length has 4800 turns and carries a current of 2.0 A. Part A Calculate the magnetic f
lux through the circular cross-sectional area of the solenoid. Hint: Assume this is a very long solenoid and use the simplified magnetic field formula for an infinite solenoid.
1 answer:
Answer: Ø = 0.034Wb
Explanation:
A solenoid is idealized as a certain number of current loops of diameter d in series.
• The current loops create a magnetic field. This field permeates each loop, ie, there is magnetic flux through each loop.
Determine the number of turns from the length and given winding rate.
• Calculate the strength of the magnetic field produced by the solenoid, then the flux through one turn (winding).
Please find the attached file for the solution.
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Answer:
mark two options:
B. TSN and ISW
D. ISN and TSW
Explanation:
Recall the definition of vertical angles as: <em>those angles opposed by the vertex, and which are the result of two lines that intersect</em>.
In the figure given there are two pairs such angles, one pair I depicted in red, and the other one in blue, while the intersecting lines are shown in green. Notice that when two lines intersect, there are always two pairs of vertical angles being generated.
Therefore, according to the notation given, two boxes need to be checked in the set of options:
1) the one corresponding to : TSN and ISW (blue ones in the attached image)
and
2) the one corresponding to ISN and TSW (red ones in the attached image)
