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

We have a particle of mass $$m$$ and charge $$q$$. Its initial velocity is purely along the $$y$$-axis, so we may write it as $$\vec v(0)=v_0\,\hat j$$. An electric field $$\vec E$$ and a magnetic field $$\vec B$$ are both directed along the $$x$$-axis; in symbols we write $$\vec E=E_0\,\hat i$$ and $$\vec B=B_0\,\hat i$$.

The force acting on a charged particle in simultaneous electric and magnetic fields is given by the Lorentz formula

$$\vec F = q\,(\vec E + \vec v \times \vec B).$$

Dividing by the mass, we obtain the acceleration

$$\frac{d\vec v}{dt}= \frac{q}{m}\,\bigl(\vec E + \vec v \times \vec B\bigr).$$

Let us resolve every quantity into Cartesian components. We denote the velocity as $$\vec v=(v_x,\,v_y,\,v_z).$$ The electric field has only an $$x$$-component $$E_0$$, while the magnetic field has only an $$x$$-component $$B_0$$. Therefore

$$\vec E=(E_0,\,0,\,0), \qquad \vec B=(B_0,\,0,\,0).$$

The magnetic part of the force requires the cross-product $$\vec v \times \vec B$$. Writing out the determinant

$$ \vec v \times \vec B= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] v_x & v_y & v_z\\[2pt] B_0 & 0 & 0 \end{vmatrix} = \hat i\,(v_y\cdot 0-v_z\cdot 0)\;-\; \hat j\,(v_x\cdot 0-v_z\cdot B_0)\;+\; \hat k\,(v_x\cdot 0-v_y\cdot B_0). $$

Simplifying term by term we get

$$\vec v \times \vec B =(0,\;v_z B_0,\;-v_y B_0).$$

Substituting the electric and magnetic contributions into the acceleration equation gives the three component equations

$$m\,\frac{dv_x}{dt}=qE_0,$$

$$m\,\frac{dv_y}{dt}=q\,v_z B_0,$$

$$m\,\frac{dv_z}{dt}=-q\,v_y B_0.$$

We now analyse each direction.

1. Along the $$x$$-axis the acceleration is constant because the right-hand side is the constant $$qE_0$$. Hence

$$\frac{dv_x}{dt}= \frac{qE_0}{m}\equiv a,$$

where we have introduced $$a=\dfrac{qE_0}{m}$$ for convenience. Integrating with the initial condition $$v_x(0)=0$$ (because the initial velocity had no $$x$$-component) we get

$$v_x(t)=a\,t=\frac{qE_0}{m}\,t.$$

2. The $$y$$- and $$z$$-components are coupled. It is helpful to introduce the cyclotron frequency

$$\omega=\frac{qB_0}{m}.$$

The two differential equations now read

$$\frac{dv_y}{dt}= \omega\,v_z,$$

$$\frac{dv_z}{dt}= -\omega\,v_y.$$

Differentiate the first of these once more with respect to time:

$$\frac{d^2 v_y}{dt^2}= \omega\,\frac{dv_z}{dt}=\omega\,(-\omega v_y)=-\omega^2\,v_y.$$

This is the standard simple-harmonic-motion equation $$\dfrac{d^2y}{dt^2}=-\omega^2 y$$ whose solution is a sinusoid. Using the initial conditions $$v_y(0)=v_0$$ and $$v_z(0)=0$$ we obtain

$$v_y(t)=v_0\cos(\omega t),$$

and, from the relation $$\dfrac{dv_y}{dt}=\omega v_z,$$

$$v_z(t)= -\,v_0\sin(\omega t).$$

We now compute the square of the speed at any instant. The speed is the magnitude of $$\vec v$$, so

$$ \begin{aligned} v^2(t)&=v_x^2(t)+v_y^2(t)+v_z^2(t)\\[4pt] &=\bigl(a t\bigr)^2+\bigl(v_0\cos\omega t\bigr)^2+\bigl(-v_0\sin\omega t\bigr)^2\\[4pt] &=a^2 t^2+v_0^2\bigl(\cos^2\omega t+\sin^2\omega t\bigr). \end{aligned} $$

Because $$\cos^2\theta+\sin^2\theta=1$$, the trigonometric terms combine neatly to unity, leaving

$$v^2(t)=v_0^2 + a^2 t^2.$$

We are asked to find the time at which the speed doubles, i.e. when $$v(t)=2v_0$$. Squaring this requirement gives

$$\bigl(2v_0\bigr)^2 = v_0^2 + a^2 t^2.$$

Simplifying step by step,

$$4v_0^2 = v_0^2 + a^2 t^2,$$

$$4v_0^2 - v_0^2 = a^2 t^2,$$

$$3v_0^2 = a^2 t^2,$$

$$t^2 = \frac{3v_0^2}{a^2}.$$

Taking the positive square root (time is positive) we have

$$t = \frac{\sqrt{3}\,v_0}{a}.$$

Finally substitute back $$a=\dfrac{qE_0}{m}$$:

$$t = \frac{\sqrt{3}\,v_0}{\dfrac{qE_0}{m}} =\frac{\sqrt{3}\,m\,v_0}{qE_0}.$$

This time corresponds exactly to Option C in the given list.

Hence, the correct answer is Option C.