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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B 2
is which shell are the valence electrons of the elements in period 2 found
Answer:
vHe / vNe = 2.24
Explanation:
To obtain the velocity of an ideal gas you must use the formula:
v = √3RT / √M
Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:
vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol
vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
The ratio is:
vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol
vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol
<em>vHe / vNe = 2.24</em>
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I hope it helps!
PH = -log10 [H+]. So anwer 2 pH
