Question

A particle of mass m and charge q has an initial velocity $$\vec{v} = v_0 \hat{j}$$. If an electric field $$\vec{E} = E_0 \hat{i}$$ and magnetic field $$\vec{B} = B_0 \hat{i}$$ act on the particle, its speed will double after a time.

Answer 1

Correct option is A. $$\dfrac{\sqrt{3} mv_0}{q E_0}$$

Given $$\vec{V}_i = V_0 \hat{j} , \ \vec{E} = E_0 \hat{i} , \vec{B} = B_0 \hat{i}$$

Now we see that the velocity is perpendicular to the magnetic field so the magnitude of velocity will not change due to magnetic field.

The magnitude of velocity will change in 'i' direction due to $$\vec{E} $$ only

$$acc = \dfrac{force}{mass} $$ due to $$\vec{E}$$ on particle $$= \dfrac{q E_0}{m} \hat{i}$$

Let us say that the speed of particle doubles after time 't'. in y direction the speed will be $$V_0$$ only in $$y - z$$ plane always perpendicular to x-direction so net speed at time 't'.

$$(V_{net})^2 = \underbrace { (V_0)^2 }_{ V \ in \ y - z \ plane } + \underbrace { \left(\dfrac{q E_0}{m} t\right)^2 }_{ V_x \ after \ time\ t \, in\ x-direction\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!} $$ .

Also by given condition $$V_{net} = 2V_0$$

$$\Rightarrow (2 V_0)^2 = V_0^2 + \left(\dfrac{q E_0}{m} t\right)^2$$

$$\Rightarrow 4 V_0^2 - V_0^2 = \left(\dfrac{q E_0}{m} t \right)^2$$

$$\Rightarrow 3 V_0^2 = \left(\dfrac{q E_0}{m} t \right)^2$$

$$\Rightarrow t = \dfrac{\sqrt{3} V_0 m}{q E_0}$$

Option (A) is correct.