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:
Explanation:
Given:
Force between pair of charges= 900 newtons
The distance between the charges = 0.01 meters
Strength of Charge first q1 = 2e-10 Coulomb
To find:
Strength of Charge second q2 = ____ Coulomb?
Solution:
We know that,
Force between two charges separate by distance r is given by the equation,
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Answer:
The speed is 1.52 m.
Explanation:
Given that,
Displacement =0.270 m
Distance = 0.130 m
Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.
We need to calculate the angular velocity
Using formula of angular velocity
Put the value into the formula
We need to calculate the velocity
Using formula of velocity
Put the value into the formula
Hence, The speed is 1.52 m.