Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.59 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2190 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.91 V/m, (b) in the negative z direction and has a magnitude of 3.91 V/m, and (c) in the positive x direction and has a magnitude of 3.91 V/m?

Answer 1

a) 1.88*10^-18 N

b) 6.32*10^-19 N

c) 1.9*10^-18 N

Step-by-step explanation:

The total force over the electron is given by:

`\vec{F}=q\vec{E}+q\vec{v}\ X\ \vec{B}`

the first term is the electric force and the second one is the magnetic force.

You have that the velocity of the electron and the magnetic field are:

`\vec{v}=2190(m)/(s)\ \hat{j}\\\\\vec{B}=-5.39*10^(-3)T\ \hat{i}`

by using the relation j X (-i) =- j X i = -(-k) = k, you obtain:

`\vec{v} \ X\ \vec{B}=(2190m/s)(3.59*10^(-3)T)\hat{k}=7.862\ T m/s`

a) For an electric field of 3.91V/m in +z direction:

`\vec{F}=q[\vec{E}+\vec{v}\ X\ \vec{B}]=(1.6*10^(-19))[3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=1.88*10^(-18)N\hat {k}\\\\F=1.88*10^(-18)N`

b) E=3.91V/m in -z direction:

`\vec{F}=(1.6*10^(-19))[-3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=6.32*10^(-19)N\hat {k}\\\\F=6.32*10^(-19)N`

c) E=3.91 V/m in +x direction:

`\vec{F}=(1.6*10^(-19))[3.91\hat{i}+7.862\ \hat{k}]N\\\\\vec{F}=[6.25*10^(-19)\ \hat{i}+1.25*10^(-18)\ \hat{k} ]N\\\\F=\sqrt{(6.25*10^(-19))^2+(1.25*10^(-18))^2}N=1.9*10^(-18)\ N`