Question: A proton travels through uniform magnetic and electric fields. The magnetic fieldis $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{5}\widehat{\mathbf{i}\mathbf{}}\mathbf{mT}$.At one instant the velocity of the proton is $\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{2000}\mathbf{}\hat{\mathbf{j}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ At that instant and in unit-vector notation, what is the net force acting on the proton if the electric field is (a) role="math" localid="1663233256112" $\mathbf{4}\mathbf{.}\mathbf{00}\hat{\mathbf{k}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$, (b) $\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}\hat{\mathbf{k}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$and (c)$\mathbf{4}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$?

$\overrightarrow{\mathrm{B}}=-2.5\hat{\mathrm{i}}\mathrm{mT}$

$\overrightarrow{\mathrm{v}}=2000\hat{\mathrm{j}}\mathrm{m}/\mathrm{s}$

The total force acting on the charged particle is the sum of the forces due to electric and magnetic fields.

Formulae are as follow:

Force acting on the charged particle due to electric field,

$F_e=qE$

Force acting on the charged particle due to magnetic field,

$F_m=qvB$

Where, $F_m$ is magnetic force, v is velocity, B is magnetic field, q is charge of particle.

The net force on the proton when $\overrightarrow{E}=4.00\hat{\mathrm{k}}\mathrm{V}/\mathrm{m}$:

$$\begin{gathered} \overrightarrow{\mathrm{F}}=\mathrm{q}\left(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}\right) \\ \overrightarrow{\mathrm{F}}=1.6\times {10}^{-19}\left(4.00\hat{\mathrm{k}}+2000\hat{\mathrm{j}}\times \left(-2.5\times {10}^{-3}\right)\hat{\mathrm{i}}\right) \\ \overrightarrow{\mathrm{F}}=1.6\times {10}^{-19}\left(4.00\hat{\mathrm{k}}+5.00\hat{\mathrm{k}}\right) \\ \overrightarrow{\mathrm{F}}=1.44\times {10}^{-18}\hat{\mathrm{k}}\mathrm{N} \end{gathered}$$

Hence, the net force acting on the proton is $\overrightarrow{F}=1.44\times {10}^{-18}\hat{k}\mathrm{N}$

Then net force on the proton when $\overrightarrow{E}=-4.00\hat{\mathrm{k}}\mathrm{V}/\mathrm{m}$:

$$\begin{gathered} \overrightarrow{\mathrm{F}}=\mathrm{q}\left(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}\right) \\ \overrightarrow{\mathrm{F}}=1.6\times {10}^{-19}\left(-4.00\hat{\mathrm{k}}+2000\hat{\mathrm{j}}\times \left(-2.5\times {10}^{-3}\right)\hat{\mathrm{i}}\right) \\ \overrightarrow{\mathrm{F}}=1.6\times {10}^{-19}\left(-4.00\hat{\mathrm{k}}+5.00\hat{\mathrm{k}}\right) \\ \overrightarrow{\mathrm{F}}=1.60\times {10}^{-19}\hat{\mathrm{k}}\mathrm{N} \end{gathered}$$

Hence, the net force acting on the proton is $\overrightarrow{\mathrm{F}}=1.60\times {10}^{-19}\hat{\mathrm{k}}\mathrm{N}$

The net force on the proton when $\overrightarrow{E}=4.00\hat{\mathrm{i}}\mathrm{V}/\mathrm{m}$

$$\begin{gathered} \overrightarrow{\mathrm{F}}=\mathrm{q}\left(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}\right) \\ \overrightarrow{\mathrm{F}}=1.6\times {10}^{-19}\left(4.00\hat{\mathrm{i}}+2000\hat{\mathrm{j}}\times \left(-2.5\times {10}^{-3}\right)\hat{\mathrm{i}}\right) \\ \overrightarrow{\mathrm{F}}=1.6\times {10}^{-19}\left(4.00\hat{\mathrm{i}}+5.00\hat{\mathrm{k}}\right) \\ \overrightarrow{\mathrm{F}}=\left(6.41\times {10}^{-19}\right)\hat{\mathrm{i}}\mathrm{N}+\left(8.01\times {10}^{-19}\right)\hat{\mathrm{k}}\mathrm{N} \end{gathered}$$

Hence, the net force acting on the proton is $\overrightarrow{\mathrm{F}}=\left(6.41\times {10}^{-19}\right)\hat{\mathrm{i}}\mathrm{N}+\left(8.01\times {10}^{-19}\right)\hat{\mathrm{k}}\mathrm{N}$.

Therefore, the values of net force due to different electric fields can be determined by taking the vector sum of forces due to the electric and magnetic field.