Question

An electron is traveling horizontally towards the east. A magnetic field in a vertically downward direction exerts a force on the electron along____

• A. East
• B. West
• C. North
• D. South

Answer 1

The correct option is C

North

Step 1: Define the forces

Given, the electron is moving horizontally towards east and magnetic field is vertically downward.
East is along the positive $x$-axis and vertically downward is along the negative $z$-axis.
From the conventions of vector notation, $\hat{i}$is the unit vector along the positive $x$-axis, and $-\hat{k}$ is the unit vector along the negative $z$-axis.

Then the velocity vector of the electron can be given as,
$\begin{array}{rcl}\overrightarrow{v}& =& a\hat{i}+0\hat{j}+0\hat{k}\end{array}$

where $a$ is the magnitude of the velocity.

Similarly, the magnetic field vector can be given as,
$\begin{array}{rcl}\overrightarrow{B}& =& 0\hat{i}+0\hat{j}-b\hat{k}\end{array}$

where $b$ is the magnitude of the magnetic field.

Step 2: Formulas used

We know that when a charged particle is moving in a magnetic field it experiences a force on it which is given as,
$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

where $q$ is the charge of the particle, $\overrightarrow{v}$ is the velocity vector of the particle and $\overrightarrow{B}$ is the magnetic field vector.

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are vectors such that $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ then their cross-product is,
$\overrightarrow{a}\times \overrightarrow{b}=\left(a_2{b}_3-b_2{a}_3\right)\hat{i}+\left(a_3{b}_1-b_3{a}_1\right)\hat{j}+\left(a_1{b}_2-b_1{a}_2\right)\hat{k}$

Step 3: Substitute values in the formula

Substituting the values of the terms in the force equation,

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$
$\begin{array}{rcl}\overrightarrow{F}& =& q\left[\left(a\hat{i}+0\hat{j}+0\hat{k}\right)\times \left(0\hat{i}+0\hat{j}-b\hat{k}\right)\right]\\ & =& q\left\{\left[0\left(-b\right)-0·0\right]\hat{i}+\left[0·0-\left(-b\right)a\right]\hat{j}+\left[a·0-0·0\right]\hat{k}\right\}\\ \therefore \overrightarrow{F}& =& q·ab\hat{j}\end{array}$

Therefore the vector of the force on the electron is $\overrightarrow{F}=q·ab\hat{j}$.

Step 4: Interpret the direction of the force

The vector part of the force is $ab\hat{j}$ and the unit vector along the direction of force is $\hat{j}$.
From the conventions of vector notation, we see that $\hat{j}$ is along the North.

Therefore the direction of force is along the North.

Hence, option C is correct.