Question

An electron is moving at $7.20\times {10}^{6}m/s$in a magnetic field of strength 83.0mT. What is the (a) maximum and (b) minimum magnitude of the force acting on the electron due to the field? (c) At one point the electron has an acceleration ofmagnitude role="math" localid="1662987933736" $4.90\times {10}^{14}m/s^2$.What is the angle between the electron’s velocity and the magnetic field?

Short answer

1. Maximum magnitude of the force acting on the electron is $F=9.56\times {10}^{-14}N$
2. Minimum magnitude of the force acting on the electron is F=0N
3. The angle between electron velocity and magnetic field is $\theta =0.{267}^{\circ }$

Step-by-step solution

Identification of given data

$$\begin{gathered} v=7.20\times {10}^{6}m/s \\ B=83\times {10}^{-3}T \\ e=1.6\times {10}^{-19}C \\ {m}_e=9.1\times {10}^{-31}kg \\ a=4.90\times {10}^{14}m/s^2 \end{gathered}$$

Understanding the concept

Magnetic force can be written from equation 28-3 as a vector product of velocity and magnetic field. From that, we can calculate largest value of force when velocity vector and magnetic field are perpendicular to each other. The smallest value of force can be calculated when velocity vector and magnetic field are both parallel to each other. After that, we can find the angle between the velocity vector and magnetic field by using Newton’s law.

Formula:

$F⃗=q(v⃗\times B⃗)$

(a) Determining the maximum magnitude of the force acting on the electron due to the field

The largest value of force occurs if the velocity and magnetic field are perpendicular to each other.

$$\begin{gathered} F=q\left(\overrightarrow{V}\times \overrightarrow{B}\right) \\ =qvB\sin \theta \end{gathered}$$

$\theta ={90}^{\circ }$

F = qvB

=$\left(1.6\times {10}^{-19}C\times 7.20\times {10}^{6}m/s\times 83\times {10}^{-3}T\right)$

$F=9.56\times {10}^{-14}N$

(b) Determining the minimum magnitude of the force acting on the electron due to the field

The smallest value of the magnetic force when the velocity and the magnetic field are parallel to each other:

$$\begin{gathered} F=q\left(\overrightarrow{V}\times \overrightarrow{B}\right) \\ =qvB\sin \theta \end{gathered}$$

$\theta =0^{\circ }$

$$\begin{gathered} F=qvB\sin 0 \\ F=0N \end{gathered}$$

(c) Determining the angle between the electron’s velocity and the magnetic field

According to Newton’s second law, F = ma

$$\begin{gathered} a⃗=F⃗/m \\ a=qvB\sin \theta /m \end{gathered}$$

By rearranging the equation, we can get

$\theta =[ma/qvB]$

$\theta =\frac{(9.1\times {10}^{-31}kg\times 4.90\times {10}^{14}m/s^2)}{(1.6\times {10}^{-19}C\times 7.20\times {10}^{6}m/s\times 83\times {10}^{-3}T)}$

$\theta =\left[\frac{(44.59\times {10}^{-17}N)}{(956.16\times {10}^{-16}N)}\right]$

$\theta =[4.66\times {10}^{-3}]$

$\theta =0.{267}^{\circ }$