Question

In a region of crossed fields as shown, the strength of electric filed $E$ and that of magnetic field is $B$. Three positively charged particles with speeds $V_1,V_2$ and $V_3$ are projected and their paths are shown. From this it implies that

• A. $V_1>\frac{E}{B}$
• B. $V_2=\frac{E}{B}$
• C. $V_3<\frac{E}{B}$
• D. All of these

Answer 1

The correct option is D All of these

Magnetc field is into the plane. Force by magnetic field, $\overrightarrow{F_B}=q(\overrightarrow{V}\times \overrightarrow{B})$
$=qvB$ (upward direction)

Direction of electric field is form positive plate to negative plate.
Force by electric field, $\overrightarrow{F_E}=q\overrightarrow{E}$
$=qE$ (downward direction)

If $\overrightarrow{F_E}=-\overrightarrow{F_B}$
Net force, $\overrightarrow{F_E}+\overrightarrow{F_B}=0$

$\therefore$ Charge particle will move in straight line. $\text{/}qE=\text{/}q{V}_{2}B$
$V_2=$$\frac{E}{B}$

But for path shown by $V_1$ means $\overrightarrow{F_B}>\overrightarrow{F_E}$

$\therefore \text{/}q{V}_{1}B>\text{/}qE$
$V_1>\frac{E}{B}$
For path shown by $V_3$ means $\overrightarrow{F_E}>\overrightarrow{F_B}$

$\therefore \text{/}qE>\text{/}q{V}_{3}B$
$V_3<\frac{E}{B}$

$\therefore$ Option $\left(D\right)$ is the correct answer.