Q141 / 155 · WBJEE 2016 · Physics
An electron enters an electric field having
intensity $\mathbf{E}=3 \overrightarrow{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \cdot \mathbf{V} \mathbf{m}^{-1}$ and magnetic
field having induction $\mathrm{B}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ T with a velocity $\mathbf{v}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ ms $^{-1}$. The magnitude of the force acting on the electron is (Given, $\left.e=-1.6 \times 10^{-19} \mathrm{C}\right)$
- A$2.02 \times 10^{-18} \mathrm{N}$
- B$5.16 \times 10^{-16} \mathrm{N}$
- C$3.72 \times 10^{-17} \mathrm{N}$
- Correct answerDNone of the above
Solution
Given, $E=3 i+6 j+2 \hat{k} \vee / m$
$$
\begin{array}{l}
B=2 \hat{i}+3 \hat{j} T \\
v=2 \hat{i}+3 \hat{j} m / s \\
e=-1.6 \times 10^{-10} C
\end{array}
$$
The magnitude of electric field
$$
\begin{aligned}
|E| &=\sqrt{(3)^{2}+(6)^{2}+(2)^{2}} \\
&=\sqrt{9+36+4} \\
&=\sqrt{49}=7
\end{aligned}
$$
The force acting on electron due to electric field
$$
F=q E
$$
$$
\begin{aligned}
|F| &=|q||\mathbf{E}| \\
&=1.6 \times 10^{-19} \times 7 \mathrm{N} \\
&=11.2 \times 10^{-19} \mathrm{N} \\
&=1.12 \times 10^{-18} \mathrm{N}
\end{aligned}
$$
Alternate Method
B $\|$ v Magnetic field is parallel to velocily of electron. $\mathrm{v} \| \mathrm{B}$ Magnetic force
$$
\begin{aligned}
F_{m} &=q(\mathbf{v} \times \mathbf{B}) \\
&=q v B \sin \theta \\
&=q v B \sin 0 \\
&=0
\end{aligned}
$$
$\therefore$ Net force $\quad F=F_{e}+F_{m}$
$$
\begin{array}{l}
=\left(1.12 \times 10^{-18}+0\right) \mathrm{N} \\
=1.12 \times 10^{-18} \mathrm{N}
\end{array}
$$
Hence, none of the given options is correct.
Hence, the correct answer is option: 'D'.