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:
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Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
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So v = 12.58 m/s or v = -15.58 m/s ( not possible)
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Answer:
The temperature change of the copper is greater than the temperature change of the water.
Explanation:
deltaQ = mc(deltaT)
Where,
delta T = change in the temperature
m =mass
c = heat capacity
The temperature change in the copper is nearly 11 times the temperature change in the water.
So, the correct option is,
The temperature change of the copper is greater than the temperature change of the water.
Hope this helps!