Path of a Charged Particle in Electric and Magnetic Field

Added: 2026-05-12

[00:00:00]This is a very interesting problem both intuitively and mathematically.

[00:00:04]We are given a particle of mass M and positive charge Q, which is released from rest at the origin.

[00:00:12]Now there is a uniform electric field E along the Y direction and a uniform magnetic field B along the Z direction.

[00:00:22]We are asked to describe the path of the particle.

[00:00:25]Now before jumping into equations, pause and try to think about what is happening physically.

[00:00:32]Two fields are acting simultaneously and both affect the motion in very different ways.

[00:00:38]The electric field tries to accelerate the charge in a straight line along its own direction.

[00:00:43]The magnetic field on the other hand tries to bend the path of the particle by applying force perpendicular to velocity.

[00:00:52]This creates something interesting like a combination of straight line and simple circular motion.

[00:00:59]Now the core law governing this motion is the Lorentz force.

[00:01:04]In simple words, the total force on a charged particle is given by the combined effect of electric plus magnetic fields.

[00:01:12]The electric force is straightforward.

[00:01:14]It is simply Q * E where E is a vector because the electric field is a vector quantity.

[00:01:22]The magnetic force is given as Q * V cross B, which is the vector cross product between velocity and magnetic field. Substitute E as EJ since the electric field is in Y direction and B as BK since the magnetic field is in the Z direction. Then we will assume V as VXI or X direction plus VYJ plus VZK and substitute that as well.

[00:01:53]Next find the cross product V cross B by multiplying each term in the first vector by the second vector. We get this. Now apply the unit vector cross product rules. I cross K is minus J, then J cross K is I and K cross K is zero. So we get VYB I minus VXBJ.

[00:02:16]Now expand this thing to get this.

[00:02:19]Combine the like terms so that we can resolve the motion along X, Y, and Z directions. So the force in X direction is this. In Y, it is this and in Z, it is zero.

[00:02:33]What can you tell about the motion of the particle in the Z direction based on the fact that the force in the Z direction is zero?

[00:02:41]Yes, right. There will be no motion in Z direction because the initial velocity of the particle is zero. Nice. So the motion remains confined to the X and Y plane. Now using calculus, we can write force in X direction as mass times acceleration in X direction, which is rate of change of VX with respect to time and that equals QB times VY. Divide by M both sides to get DVX over DT as this. Similarly, M times DVY over DT equals this. Now bring B outside to get this as QB times E over B minus VX. Then divide by M both sides to get this.

[00:03:24]Now to simplify the equations, we define a constant W equal to Q multiplied by B divided by M.

[00:03:32]This constant is also known as the cyclotron frequency. Using this, the equations become rate of change of velocity in X equals W times velocity in Y and rate of change of velocity in Y equals W times E divided by B minus velocity in X.

[00:03:53]Now how to solve these coupled equations? Here the rate of change of X is dependent upon Y and vice versa. In order to solve such equations, differentiate the second equation again with respect to time.

[00:04:08]This gives the second derivative of VY equals minus W times the rate of change of VX. But the rate of change of VX is W times VY and thus we get the second derivative of VY equals minus W squared times VY.

[00:04:27]Wow, can you observe something? This is exactly the form of simple harmonic motion, which means velocity in Y varies sinusoidally. So let us write VY as A multiplied by sine of W times time T where A is a constant we still need to find. We will use initial conditions to determine this constant. At time zero, the particle is released from rest. VY is zero.

[00:04:56]This condition is already satisfied because the sine of zero is zero.

[00:05:00]From this at T equals zero, VX is zero and thus DVY over DT is WE divided by B.

[00:05:08]Differentiate this VY to get DVY over DT equals AW times cosine WT. Put T equals zero to get DVY over DT at time zero equals AW times cosine zero, which is simply AW. So equating both of them, we get WE divided by B equals AW. And thus A equals E over B. So we get VY equals E divided by B multiplied by the sine of WT.

[00:05:38]Now put this VY here to get the rate of change of VX as this.

[00:05:43]To find VX, take DT this side and simply integrate this from time zero to T.

[00:05:49]This side, the limits will be zero to some VX because at time zero, VX is zero.

[00:05:56]So we get VX as WE over B times integration of sine WT, which is minus cosine WT over W. So we get E over B times cosine WT from T to zero. Put the values of T and we get VX as E over B times 1 minus cosine WT. Once we have velocities in X and Y direction, we can again integrate them with respect to time to find the positions X and Y of the particle considering the fact that initially the particle was at the origin.

[00:06:34]On doing the math, we get X as this and Y as this. If we use another variable R as E over WB, we get X and Y as these.

[00:06:46]Oh wow, you know what? These equations describe a cycloid, which means the path traced by a point on the circumference of a circle as it rolls along a straight line without slipping. So the path of the particle is a cycloid and it moves like this. That was super amazing. Now can you let me know in the comments what will be the value of the maximum speed that the particle can obtain? If you enjoyed this video, please don't forget to like, share, and subscribe to our channel. Also, you can support my channel by joining our community and becoming a member. So good.