When an electron passes through the magnetic field of a horseshoe magnet, the electron's direction is changed.
Path of an electron in a magnetic field
The force (F) on wire of length L carrying a current I in a magnetic field of strength B is given by the equation:
F = BIL
But Q = It and since Q = e for an electron and v = L/t you can show that :
Magnetic force on an electron = BIL = B[e/t][vt] = Bev where v is the electron velocity
In a magnetic field the force is always at right angles to the motion of the electron (Fleming's left hand rule) and so the resulting path of the electron is circular.
Therefore :
Magnetic force = Bev = mv2/r = centripetal force
v = [Ber]/m
and so you can see from these equations that as the electron slows down the radius of its orbit decreases.
If the electron enters the field at an angle to the field direction the resulting path of the electron (or indeed any charged particle) will be helical. Such motion occurs above the poles of the Earth where charges particles from the Sun spiral through the Earth's field to produce the aurorae.
To learn more about electron : brainly.com/question/860094
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The correct answer to this question is the unique atomic number which would be B
Answer:
This is because of scintillation ("Twinkling") as the light passes through the atmosphere of the Earth. As the air moves in and out, the starlight is refracted, often different colors in different directions. Because of this "chromatic abberation," stars can appear to change colors when they are twinkling strongly.
Explanation:
I hope that you can understand this!!! Lol
The thing that you have to pull back to release with, that would be considered a third class lever.
I hope this helps. :)
Answer:
The mass defect of a deuterium nucleus is 0.001848 amu.
Explanation:
The deuterium is:
The mass defect can be calculated by using the following equation:
![\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20%5BZm_%7Bp%7D%20%2B%20%28A%20-%20Z%29m_%7Bn%7D%5D%20-%20m_%7Ba%7D)
Where:
Z: is the number of protons = 1
A: is the mass number = 2
: is the proton's mass = 1.00728 amu
: is the neutron's mass = 1.00867 amu
: is the mass of deuterium = 2.01410178 amu
Then, the mass defect is:
![\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20%5B1.00728%20amu%20%2B%20%282-%201%291.00867%20amu%5D%20-%202.01410178%20amu%20%3D%200.001848%20amu)
Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.
I hope it helps you!
