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Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thousands of miles from Earth, they interact with Earth's magnetic field of magnitude 3.04 ✕ 10−8 T in the positive z-direction. Find the magnitude and direction of the magnetic force on a proton. Find the magnitude and direction of the magnetic force on an electron.

Answer :

okpalawalter8

Answer:

[tex]F=2.84*10^{-26}N[/tex]  & -y direction

[tex]F=2.84*10^{-26}N[/tex] & +y direction

Explanation:

From the question we are told that:

Speed of electron [tex]V_e=3.83 * 10^5 m/s[/tex] +x direction

Earths magnetic field [tex]B_e=3.04 * 10^-^8[/tex] +z direction

a)

Generally the equation for magnetic force [tex]F_m[/tex] is mathematically given by

[tex]F=q(V_e*B_e)[/tex]

where

[tex]q=1.6*10^{-19}c\\\=i*\=z=-\=j[/tex]

[tex]F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)[/tex]

[tex]F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)[/tex]

[tex]F=-2.84*10^{-26}N \=j[/tex]

Magnitude & Direction

[tex]F=2.84*10^{-26}N[/tex]  -y direction

b)

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

[tex]\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)[/tex]

[tex]\=F'=-2.84*10^{-26}N \=j[/tex]

Magnitude & Direction

[tex]F=2.84*10^{-26}N[/tex] & +y direction

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