Question

An electron moves in the magnetic field$\stackrel{\rightharpoonup }{B}=0.50\hat{i}T$ with a speed of $1.0\times {10}^{7}m/s$in the directions shown in FIGURE EX29.27. For each, what is magnetic force on the electron? Give your answers in component form.

Short answer

a) The magnetic force on the electron for the component a is determined as $\overrightarrow{F}=8\times {10}^{-13}\hat{j}N$

b)The magnetic force on the electron for the component b is determined as $\overrightarrow{F}=(0.56\times {10}^{-12}\widehat{-j}+0.56\times {10}^{12}\widehat{-k})N$

Step-by-step solution

Introduction

The given is the speed of the electron $1.0\times {10}^{7}m/s$moves in the magnetic field of $\overrightarrow{B}=0.50\hat{l}T$

The objective is to find the magnetic force on the electron for each component

The method used here is the Ampere's Experiment to find the magnetic force

Step 1

a)

The magnetic field imposes a magnetic force on the moving charge, which is dependent on the velocity vector, as demonstrated by Ampere's experiment. The force increases as the angle between the velocity and the magnetic field increases.

The magnetic force can be calculated using an equation of the type

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ $\left(1\right)$

The charge is denoted by localid="1648902125297" $\mathrm{q}$.The angle between the velocity and the magnetic field is $90°$but the magnetic field's direction is negative $\mathrm{z}$hence the velocity vector is

$\overrightarrow{v}=(-1\times {10}^{7}m/s)\hat{k}$

Step 2

We plug the values for $\overrightarrow{B},qand\overrightarrow{v}$into equation$\left(1\right)$to get $\overrightarrow{F}$, where $(\hat{k}\times \hat{i})=\hat{j}$is the magnetic field component $\overrightarrow{B}=0.50\hat{i}T$.

$$\begin{gathered} \overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B} \\ =(-1.6\times {10}^{-19}C[(-1\times {10}^{7}m/s)\hat{k}\times (0.50\hat{i}T)] \\ =(-1.6\times {10}^{-19}C[-0.5\times {10}^7(\hat{k}\times \hat{i}\left)\right] \\ =\boxed{\mathit{8}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{13}\mathit{}}\hat{\mathit{j}}\mathit{}\mathit{N}} \end{gathered}$$

Step 3

b)

In the $-yz$plane, the angle between the velocity and the magnetic field is ${45}^{°}$, hence the velocity vector is

$\overrightarrow{v}=(-1\times {10}^{7}m/s)\cos {45}^{°}\hat{j}+(1\times {10}^{7}m/s)\sin {45}^{°}\hat{k}$

We plug the values for localid="1648902389473" $\overrightarrow{B},qand\overrightarrow{v}$into equation localid="1648902410608" $\left(1\right)$to get localid="1648902415023" $\overrightarrow{F}$, where localid="1648902419896" $(\hat{j}\times \hat{i})$is the magnetic field component.

localid="1648902430763" $$\begin{gathered} \overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B} \\ =(1.6\times {10}^{-19}C)\left[\right(-1\times {10}^{7}m/s)\cos {45}^{°}\hat{j}+(1\times {10}^{7}m/s)\sin {45}^{°}\hat{k}\times (0.50\hat{i}T) \\ =\boxed{\mathit{(}\mathit{0}\mathit{.}\mathit{56}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{12}}\widehat{\mathit{-}\mathit{j}}\mathit{+}\mathit{0}\mathit{.}\mathit{56}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{12}}\widehat{\mathit{-}\mathit{k}}\mathit{)}\mathit{N}} \end{gathered}$$