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.
You might be interested in
Answer:
K = 80.75 MeV
Explanation:
To calculate the kinetic energy of the antiproton we need to use conservation of energy:
<em>where : is the photon energy,
: are the rest energies of the proton and the antiproton, respectively, equals to m₀c²,
: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>
Therefore the kinetic energy of the antiproton is:
<u>The proton mass is equal to the antiproton mass, so</u>:
Hence, the kinetic energy of the antiproton is 80.75 MeV.
I hope it helps you!