Question

An electron experiences a magnetic force of magnitude 4.60 x 10-15 N when moving at an angle of 60.0 degrees with respect to a magnetic field of magnitude 3.50 x 10-3 T. Find the speed of the electron.

Answer 1

Concepts and reason

The concepts used to solve this problem are magnetic force, magnetic field, and charge of one electron.

First find the speed of the electron by using the Lorentz magnetic force exerted on the electron by substituting magnetic field, charge of an electron and angle between the field and velocity vector. Then, rearrange the equation for velocity.

Fundamentals

In terms of vector expression, the magnetic Lorentz force can be equated as,

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

Here, F is the Lorentz magnetic force, B is the magnetic field, q is the charge of an electron, and v is the speed of the electron.

The magnetic force between the two objects depends on the charge and its motion for the two objects, and the distance between them. The direction of the force depends on the direction of motion of the charge.

The expression for the magnetic force of the electron is,

$F = Bqv\sin \theta$

Here, $\theta$ is the angle between the field and velocity vector.

In terms of vector, the expression for magnetic Lorentz force is given by,

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

The vector product of the expression of magnetic force can be changed into scalar product by using angle between field vector and velocity vector.

The expression for the magnetic force of the electron is,

$F = Bqv\sin \theta$

Rewrite the expression in terms of v.

$v = \frac{F}{{Bq\sin \theta }}$

The expression for the speed of electron in terms of magnetic field is,

$v = \frac{F}{{Bq\sin \theta }}$

Substitute $4.60 \times {10^{ - 15}}\,{\rm{N}}$ for F, $60.0^\circ$ for $\theta$, $3.50 \times {10^{ - 3}}\,{\rm{T}}$ for B, and $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $q$ in the above equation $v = \frac{F}{{Bq\sin \theta }}$ to calculate v.

$\begin{array}{c}\\v = \frac{{\left( {4.60 \times {{10}^{ - 15}}\,{\rm{N}}} \right)}}{{\left( {3.50 \times {{10}^{ - 3}}\,{\rm{T}}} \right)\left( {1.602 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\sin 60.0^\circ }}\\\\ = 0.9473 \times {10^7}\,{\rm{m/s}}\\\\ = {\rm{9}}{\rm{.473}} \times {10^6}\,{\rm{m/s}}\\\end{array}$

Ans:

The speed of the electron is ${\rm{9}}{\rm{.473}} \times {10^6}\,{\rm{m/s}}$.