Chapter 28, Problem 1P

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming that the electron has an instantaneous velocity of 6.00 × 10⁶ m/s directed to the east.

To determine

The gravitational, electric and magnetic force on an electron

Answer to Problem 1P

The gravitational, electric and magnetic force on an electron is $8.93\times {10}^{-30}\text{N}$, $1.60\times {10}^{-17}\text{N}$ upwards and $4.80\times {10}^{-17}\text{N}$ downwards.

Explanation of Solution

Given info: The magnetic field is $50\text{μT}$ northward and electric field is $100\text{N}/\text{C}$ downward, velocity of electron is $6\times {10}^6\text{m}/\text{s}$

Explanation:

The formula to calculate the gravitational force is,

$F=mg$

Here,

$m$ is the mass of the particle.

$g$ is the acceleration due to gravity.

Substitute $9.11\times {10}^{-31}\text{Kg}$ for $m$ and $9.8\text{m}/{\text{s}}^2$ for $g$.

$\begin{array}{c}F=mg\\ =\left(9.11\times {10}^{-31}\text{Kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =8.93\times {10}^{-30}\text{N}\end{array}$

The formula to calculate the electric force is,

$F=qE$

Here,

$q$ is the charge of electron.

$E$ is the electric field.

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$ and $100\text{N}/\text{C}$ for $E$

$\begin{array}{c}F=qE\\ =\left(1.6\times {10}^{-19}\text{C}\right)\left(100\text{N}/\text{C}\right)\\ =1.60\times {10}^{-17}\text{N}\end{array}$

The direction of electric force is upwards.

The formula to calculate the magnetic force is,

$F=qvB$

Here,

$q$ is the charge of electron.

$v$ is the velocity of electron.

$B$ is the magnetic field.

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $50\text{μT}$ for $B$, $6\times {10}^6\text{m}/\text{s}$ for $v$ in above expression.

$\begin{array}{c}F=qvB\\ =\left(1.6\times {10}^{-19}\text{C}\right)\left(50\text{μT}\right)\left(6\times {10}^6\text{m}/\text{s}\right)\\ =4.80\times {10}^{-17}\text{N}\end{array}$

The direction of magnetic force is downwards.

Conclusion:

Therefore, the gravitational, electric and magnetic force on an electron is $8.93\times {10}^{-30}\text{N}$, $1.60\times {10}^{-17}\text{N}$ upwards and $4.80\times {10}^{-17}\text{N}$ downwards.