Question

Determine the magnitude and direction of the force on an electron traveling 8.75 x 10⁵m/s horizontally to the east in a vertically upward magnetic field of strength 0.45 T.

I got that v= v j <----vector and B= B k <----vector, but how do i know this???

The answer is

F= (-e) v x B

F= -e v B i <-----why i is positive when in the diagram above k x j = -i ???

F= -1.6 x 10^-19 x 8.75 x 10⁵ x 0.45 j N <---- why is i the j vector know???

This is really confusing please help

                 

Answer 1

Concepts and reason

The concepts used to solve this problem are a Lorentz force and right-hand rule.

Use the expression Lorentz force to determine the magnitude and direction of the force of the particular points by substituting the provided value of that point.

The direction of the force on an electron traveling horizontally to the east in a vertically upward magnetic field.

Fundamentals

Write the expression for the Lorentz force.

$\begin{array}{c}\\\vec F = q\left( {\vec v \times \vec B} \right)\\\\ = qvB\left( {\sin \theta } \right)\\\end{array}$

Here, $F$is the force exerted, $B$is a magnetic field, $q$ is the charge and $v$is the velocity of the electron.

Calculate maximum force.

$\vec F = qvB\left( {\sin \theta } \right)$

Substitute $90^\circ$ for $\theta$.

$\begin{array}{c}\\F = qvB\left( {\sin 90^\circ } \right)\\\\ = qvB\\\end{array}$

Substitute $0.45{\rm{ T}}$for $B$, $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $q$ and $8.75 \times {10^5}{\rm{ m/s}}$ for $v$.

$\begin{array}{c}\\F = \left( {0.45{\rm{ T}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {8.75 \times {{10}^5}{\rm{ m/s}}} \right)\\\\ = 6.3 \times {10^{ - 14}}{\rm{ N}}\\\end{array}$

The direction of the force exerted can be determined from the right-hand rule.

From the expression of the force, it is clear that if the velocity is known and the direction of the magnetic field is known then from cross vector product the direction of the force is also known.

$\vec F = q\left( {\vec v \times \vec B} \right)$

From the provided question, the direction of the velocity is eastward and the direction of the magnetic field is upward then from vector cross product the direction of the force is upward.

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Ans:

The magnitude and direction of the force on an electron is $6.3 \times {10^{ - 14}}{\rm{ N}}$ towards North.