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.

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

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.

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" $F_B=-0.624\times {10}^{-13}N\hat{k}~\left(6.2\times {10}^{-14}N\right)\hat{k}$

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.