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Step-by-Step Solution
Step 1: Identify the forces acting on the particle
The particle has charge $q$, mass $m$, and an initial velocity $\overrightarrow{v_0} = v_0 \,\hat{j}$. It moves in the presence of an electric field $\overrightarrow{E} = E_0 \,\hat{i}$ and a magnetic field $\overrightarrow{B} = B_0 \,\hat{i}.$
The electric force on the particle is $\overrightarrow{F}_\text{electric} = q\overrightarrow{E} = q E_0 \,\hat{i}.$
The magnetic force is given by $\overrightarrow{F}_\text{magnetic} = q (\overrightarrow{v} \times \overrightarrow{B})$. Since $\overrightarrow{v}$ is initially along $\hat{j}$ and $\overrightarrow{B}$ is along $\hat{i}$, the magnetic force is perpendicular to both $\hat{i}$ and $\hat{j}$. This magnetic force does not do any work on the particle (it changes direction but not the speed). Hence, it does not affect the magnitude of the velocity.
Step 2: Determine acceleration due to the electric field
The net change in speed is influenced by the electric field alone. The force in the $x$-direction due to the electric field is:
$F_x = q E_0 \quad \Rightarrow \quad a_x = \dfrac{F_x}{m} = \dfrac{q E_0}{m}.$
Because of this constant acceleration in the $x$-direction, the $x$-component of velocity at time $t$ is
$v_x(t) = a_x \, t = \dfrac{q E_0}{m}\,t.$
Step 3: Express the total speed at time t
Initially, the velocity was purely in the $y$-direction with magnitude $v_0$. After time $t$, the velocity has two components:
$v_x = \dfrac{q E_0}{m} t, \quad v_y = v_0.$
Hence, the total speed $v$ is given by
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{\Bigl(\dfrac{q E_0}{m}\,t\Bigr)^2 + v_0^2}.$
Step 4: Impose the condition that the speed doubles
We want the speed to become $2v_0$. Therefore, set $v = 2v_0$:
$ \sqrt{\Bigl(\dfrac{q E_0}{m} t\Bigr)^2 + v_0^2} = 2\,v_0.$
Square both sides:
$\Bigl(\dfrac{q E_0}{m} t\Bigr)^2 + v_0^2 = 4\,v_0^2.$
Rearrange to solve for $t$:
$\Bigl(\dfrac{q E_0}{m} t\Bigr)^2 = 4\,v_0^2 - v_0^2 = 3\,v_0^2.$
Thus,
$\dfrac{q E_0}{m} \, t = \sqrt{3}\,v_0 \quad \Rightarrow \quad t = \dfrac{\sqrt{3}\,m\,v_0}{q\,E_0}.$
Step 5: Final result
Therefore, the time required for the particle's speed to double is
$t = \dfrac{\sqrt{3}\,m\,v_0}{q\,E_0}.$
