What will be the path of this particle in combined electric and magnetic fields?

Question

Say a particle is moving with constant velocity along the positive y-axis. Both electric and magnetic fields are applied along the positive x-axis. What will be the path of this particle?

I am confused between these two solutions:

(1) Because the particle is moving along the y-axis and magnetic field is along x-axis, the magnetic force must be directed along the negative z-axis while electric force will be along positive x-axis. The resultant force will be in x-z plane which is perpendicular to the direction of velocity and hence the particle will execute circular motion with constant speed.

(2) The magnetic force will be in negative z-axis and electric force will be along positive x-axis. The magnetic force will make the particle execute circular motion in the y-z plane and the electric force will take the particle in the positive x direction. The combined motion will create a helix.

Answer 1

It's Option 2. To see this, write down the differential equations for the motion. We have Newton's Second Law as $m \dot{\vec{v}} = q (\vec{E} + \vec{v} \times \vec{B})$; assuming that both fields point in the $x$-direction ($\vec{E} = E \hat{\imath}$ and $\vec{B} = B \hat{\imath}$), the components of the Newton's Second Law are \begin{align} m \dot{v}_x &= q E\\ m \dot{v}_y &= q B v_z \\ m \dot{v}_z &= - q B v_y \end{align} You may or may not be able to solve these equations immediately. But what we can see is that the motion in the $x$-direction is completely independent of the motion in the $yz$-plane; the equation for $v_x$ doesn't include $v_y$ or $v_z$, and the equations for $v_y$ and $v_z$ don't involve $v_x$.

Moreover, the equation for $v_x$ is just what it would be for a particle in an electric field (without a magnetic field), and the equations for $v_y$ and $v_z$ are just what they would be for a charged particle in a magnetic field (without an electric field.) So the particle executes uniformly accelerated motion along the $x$-axis, and circular motion parallel to the $yz$-plane. The result is a helix whose pitch increases along the path of the particle.

^{Footnote: This all assumes that the charged particle is non-relativistic. If the speed of the particle becomes comparable to the speed of light, then the analysis becomes more complicated; but the basic conclusion, that the path is a helix of increasing pitch, remains essentially the same.}

Answer 2

Assuming that particle as a charge ,force acting on a charge $q$ is

$$ F = q \left( v \times B \right) \tag{1} \label{1} $$

So let say $q$ is along $+y$ axis and $E$ and $B$ (both the fields) along $+x$ axis. Fine use right hand screw rule (curl your right hand from direction of $v$ to $B$ and thumb points the direction of force) along $v$ to $B$ (that is $v \times B$) and we see see the thumb point along $- Z$ .

Now we also can see that the force is perpendicular to velocity of particle.

Power, if you remember is $F \cdot v$ or $F v \cos \left( \theta \right)$. $\theta = 90^{o}$ so power $= 0$ and work also as power = work/time. So, the particle will perform circular motion.

Here's why 2nd case won't happen. The helical motion is when velocity and electric field makes an acute angle but our $\theta$ here is $90^{o}$. If it's less than $90$, motion goes helical.