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

Answer:

$F=2.84*10^{-26}N$  & -y direction

$F=2.84*10^{-26}N$ & +y direction

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