Magnetic force on a moving charge particle with friction force acting on it

[siddscool19]

Homework Statement

A "q" charge particle is moving along +y axis with velocity V₀ starting from Origin. A friction force is acting on the charge particle " f= -(alpha)V(Vector) ".A constant magnetic field of magnitude B₀ acting along +Z axis. The mass of the particle is m. Find the x co-ordinate of the particle when it stops.

Homework Equations

The Attempt at a Solution

I Have found that the total path moved by the particle is mV₀/(alpha)

But I don't know how to find x coordinate with it.

 

[hikaru1221]

There is a quick way to solve this problem: write down the Newton 2nd law equation in vector form and integrate it

 

[IEVaibhov]

I have solved the problem. I am just too lazy to type the WHOLE thing out here. The solution is kind of long. Basically, get the general velocity along X axis and Y axis as a function of time.
To obtain this, in the penultimate step, you will get a differential equation that you will have to solve using Complex numbers.
In the end, the velocities will be something of the type v0*[e(^(-constant))]*cos(wt), where w is the angular frequency.

 

[cupid.callin]

I have solved the problem. I am just too lazy to type the WHOLE thing out here.

You are not even allowed to write whole solution. you can just give hints.

 

[rwisz]

I would approach this problem by first considering the forces acting on the charge particle. The magnetic force, given by F = qV x B, will cause the particle to move in a circular path in the xy-plane. However, the friction force, given by f = -(alpha)V, will act in the opposite direction and slow down the particle's motion.

To find the x-coordinate when the particle stops, we need to find the point at which the forces balance each other out and the particle comes to a rest. This can be done by setting the net force equal to zero and solving for the velocity at that point.

F_net = F_magnetic + F_friction = 0

qV x B + (-alpha)V = 0

Solving for V, we get V = (alpha/qB).

Now, we can use the equation for velocity, V = dx/dt, to find the x-coordinate when the particle stops.

dx/dt = (alpha/qB)

Integrating both sides, we get x = (alpha/qB)t + C

To determine the constant C, we can use the initial conditions given in the problem. Since the particle starts at the origin, we know that x = 0 when t = 0. Therefore, C = 0.

Thus, the final equation for the x-coordinate when the particle stops is x = (alpha/qB)t.

We can now substitute in the given values to find the specific x-coordinate.

x = (alpha/qB)(mV0/alpha) = mV0/qB

Therefore, the x-coordinate of the particle when it stops is mV0/qB.

In conclusion, by considering the forces acting on the charge particle and using the equation for velocity, we were able to determine the x-coordinate when the particle comes to a rest. This solution can be verified by plugging in the values and checking that the net force is indeed equal to zero.