Question

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 1

`F=2.84*10^(-26)N` & -y direction

`F=2.84*10^(-26)N` & +y direction

Step-by-step explanation:

From the question we are told that:

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

Earths magnetic field
`B_e=3.04 * 10^-^8` +z direction

a)

Generally the equation for magnetic force
`F_m` is mathematically given by

`F=q(V_e*B_e)`

where

`q=1.6*10^(-19)c\\\=i*\=z=-\=j`

`F=1.6*10^(-19)(3.83 * 10^5 m/s*3.04 * 10^-^8)`

`F=1.6*10^(-19)(3.83 * 10^5 m/s*3.04 * 10^-^8)`

`F=-2.84*10^(-26)N \=j`

Magnitude & Direction

`F=2.84*10^(-26)N` -y direction

b)

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

`\=F'=-1.6*10^(-19)(3.83 * 10^5 m/s*3.04 * 10^-^8)`

`\=F'=-2.84*10^(-26)N \=j`

Magnitude & Direction

`F=2.84*10^(-26)N` & +y direction