Question

An electron that has velocity $\overrightarrow{\mathbf{\nu }}$=($\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$m/s)$\hat{\mathbf{i}}$+ (3.0$\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$m/s)moves through the uniform magnetic field $\overrightarrow{\mathbf{B}}$= $\left(0.030T\right)\hat{\mathbf{i}}\mathbf{}\mathbf{-}\left(0.15T\right)\hat{\mathbf{j}}$.(a)Find the force on the electron due to the magnetic field. (b)Repeat your calculation for a proton having the same velocity.

Short answer

$\hat{j}$a. Magnetic force experienced by the electron is$F_B=0.624\times {10}^{-13}N\hat{k}~\left(6.2\times {10}^{-14}N\right)\hat{k}$

b. Magnetic force experienced by the proton is$F_B=-0.624\times {10}^{-13}N\overline{k}$$~6.2\times {10}^{-14}N\hat{k}$

Step-by-step solution

Given 

$$\begin{gathered} \overrightarrow{v}=\left(2.0\times {10}^6\right)\overline{i}+\left(3.0\times {10}^6\right)\hat{j} \\ \overrightarrow{B}=\left(0.030\right)\hat{i}-\left(0.15\right)\hat{j} \end{gathered}$$

Determining the concept 

Find the magnetic force on the electron using the formula for magnetic force in terms of charge, magnetic field strength, and velocity of electron.

Formulae are as follow:

$\overrightarrow{F_B}=e\overrightarrow{v}\times \overrightarrow{B}$

Where, F_B is magnetic force, v is velocity, B is magnetic field, e is charge of particle.

(a) Determining the magnetic force experienced by the electron 

The magnetic force experienced by the electron is,

$\overrightarrow{F_B}=e\overrightarrow{v}\times \overrightarrow{B}$

$F_B=\left(-1.6\times {10}^{-19}\right)\left[\left(\left(2.0\times {10}^6\right)\hat{i}+\left(3.0\times {10}^6\right)\hat{j}\right)\times \left(\left(0.030\right)\hat{i}-\left(0.15\right)\hat{j}\right)\right]$

$$\begin{gathered} \overrightarrow{v}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2.0\times {10}^6& 3.0\times {10}^6& 0\\ 0.030& -0.15& 0\end{array}\right| \\ \overrightarrow{v}\times \overrightarrow{B}=\left[\left(\left(2.0\times {10}^6\right)\left(-0.15\right)\right)-\left(3.0\times {10}^6\right)\left(0.030\right)\right]\hat{k} \\ \overrightarrow{v}\times \overrightarrow{B}=\left(-0.39\times {10}^6\right)\hat{k}\frac{m}{s}T \end{gathered}$$

Hence,

$$\begin{gathered} F_B=\left(-1.6\times {10}^{-19}\right)\left(-0.39\times {10}^6\right)\hat{k} \\ {F}_B=0.624\times {10}^{-13}N\hat{k}~\left(6.2\times {10}^{-14}N\right)\hat{k} \end{gathered}$$

Hence, the magnetic force experienced by the electron islocalid="1663047307015" $F_B=-0.624\times {10}^{-13}N\hat{k}~\left(6.2\times {10}^{-14}N\right)\hat{k}$

(b) Determining the magnetic force experienced by the proton

Since, proton will experience the same magnetic field, and its velocity is the same but charge on it is positive.

Therefore, the magnetic force experienced by proton is $F_B=-0.624\times {10}^{-13}N\hat{k}~-6.2\times {10}^{-14}N\hat{k}$

Hence, the magnetic force experienced by the proton is $F_B=-0.624\times {10}^{-13}N\hat{k}~-6.2\times {10}^{-14}N\hat{k}$

Therefore, the magnetic force on the electron and proton can be found using the formula for magnetic force in terms of charge, magnetic field strength, and velocity of electron.