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

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.

• Related questions; physics.stackexchange.com/q/234315/357431 and physics.stackexchange.com/q/758658/357431 Commented Sep 26, 2023 at 16:06
• Just write the differential equations for the vector r and solve it. If not familiar with differential equation take small time steps, E only accelerates constant in x direction, while B depends on the direction of v and its magnitude. Commented Sep 26, 2023 at 17:20

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.

• Your solution seems right considering the fact that the problem from where I took this case can be solved considering the helical case yet I am not convinced for I am not familiar with the usage of diff. equations in physics. But can you tell me why my first analysis is wrong? Why cannot I take the resultant force and then analyse the situation? Anyways Thank You!! Commented Oct 6, 2023 at 15:21
• @PrasoonJha: The presence of the electric field means that there will always be a component of the force parallel to the $x$-axis. So as soon as the particle starts moving in the $x$-direction, the force will no longer be perpendicular to the velocity, and your analysis for Option 1 fails. Commented Oct 6, 2023 at 15:39
• Thank you so much sir, I got it. Commented Oct 8, 2023 at 3:37

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.

• Thank you so much for trying to help me out. Though I think the solution by @Michael Seifert is correct. Commented Oct 8, 2023 at 3:40