91Ó°ÊÓ

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

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

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

$\stackrel{→}{B}=$$\left(400\mathrm{μ}T\right)\hat{i}$

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

Find the value ofthe electric field$\stackrel{→}{\mathbf{E}}$ 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-

$\stackrel{→}{F}=e\left(\stackrel{→}{E}+\left(\stackrel{→}{v}×\stackrel{→}{B}\right)\right)$

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

$\stackrel{→}{F}=m\stackrel{→}{a}$

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,

$\stackrel{→}{F}=e\left(\stackrel{→}{E}+\left(\stackrel{→}{V}×\stackrel{→}{B}\right)\right)$

But, according to Newton’s second law,

$\stackrel{→}{F}=m\stackrel{→}{a}$

Hence,

$m\stackrel{→}{a}=e\left(\stackrel{→}{E}+\left(\stackrel{→}{v}×\stackrel{→}{B}\right)\right)$

$\stackrel{→}{E}=\frac{1}{e}m\stackrel{→}{a}-\left(\stackrel{→}{v}×\stackrel{→}{B}\right)····················\left(1\right)$

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

Hence,

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

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

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