An electron has an initial velocity of ($\mathbf{12}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{15}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$) km/s and a constant acceleration of ($\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{12}}$ m/s²)$\hat{i}$in a region in which uniform electric and magnetic fields are present. If localid="1663949206341" $\overrightarrow{\mathbf{B}}\mathbf{=}\left(400\mathrm{\mu T}\right)\hat{\mathbf{i}}$find the electric fieldlocalid="1663949212501" $\overrightarrow{\mathbf{E}}$.

$\overrightarrow{v}=$$\left(\left(12.0\right)\hat{j}+\left(15.0\right)\hat{k}\right)km/s$

$=\left(\left(12.0\times {10}^3\right)\hat{j}+\left(15.0\times {10}^3\right)\hat{k}\right)m/s$

$\overrightarrow{a}=$$\left(2.00\times {10}^{12}m/s^2\right)\hat{i}$

$\overrightarrow{B}=$$\left(400\mu T\right)\hat{i}$

role="math" localid="1662369878702" $=\left(400\times {10}^{-6}T\right)\hat{i}$

Find the value ofthe electric field$\overrightarrow{\mathbf{E}}$ by equatingtheelectromagnetic force withtheforce given by Newton’s second law.

Newton's second law states that the time rate of change of the momentum of a body gives the force imposed on it.

Force acting on the charge in the presence of both magnetic and electric field is-

$\overrightarrow{F}=e\left(\overrightarrow{E}+\left(\overrightarrow{v}\times \overrightarrow{B}\right)\right)$

The force on the particle according to newton’s law is-

$\overrightarrow{F}=m\overrightarrow{a}$

Where, F is force, v is velocity, m is mass, E is electric field, B is magnetic field, e is charge of particle, a is acceleration.

The net force experienced by an electron is,

$\overrightarrow{F}=e\left(\overrightarrow{E}+\left(\overrightarrow{V}\times \overrightarrow{B}\right)\right)$

But, according to Newton’s second law,

$\overrightarrow{F}=m\overrightarrow{a}$

Hence,

$m\overrightarrow{a}=e\left(\overrightarrow{E}+\left(\overrightarrow{v}\times \overrightarrow{B}\right)\right)$

$\overrightarrow{E}=\frac{1}{e}m\overrightarrow{a}-\left(\overrightarrow{v}\times \overrightarrow{B}\right)····················\left(1\right)$

$\overrightarrow{v}\times \overrightarrow{B}=$role="math" localid="1662371476135" $\left|\begin{array}{c}\hat{i}\\ 0\\ 400\times {10}^{-6}T\end{array}\begin{array}{c}\hat{j}\\ 12.0\times {10}^{3}m/s\\ 0\end{array}\begin{array}{c}\hat{k}\\ 15.0\times {10}^{3}m/s\\ 0\end{array}\right|$

Hence,

$\overrightarrow{E}=$$\left[\frac{9.1\times {10}^{-31}\left(2.00\times {10}^{12}\hat{i}\right)}{\left(-1.6\times {10}^{-19}\right)}-\left[\left(6\right)\hat{j}+\left(-4.8\right)\hat{k}\right]\right]V/m$

$\overrightarrow{E}=$$\left[-11.4\hat{i}-6\hat{j}+4.8\hat{k}\right]V/m$

Hence, The electric field $\overrightarrow{E}$is $\left[-11.4\hat{i}-6.00\hat{j}+4.80\hat{k}\right]V/m$.