Calculating the Lorentz Force of an Electron in Magnetic Field

[willydavidjr]

A diagram I attached (you can view the website) shows a uniform magnetic field of magnetic flux density B[Wb/m^2] parallel to the x-axis. On the xy-plane, an electron of electric charge -e[C] and mass m[kg] departed from the x-axis in the direction of \theta[rad] with a speed of v[m/s]. Then the electron carried out a spiral movement on a cylinder whose side contains the x-axis.

This is the question: What is the magnitude of the Lorentz force acting on the electron?

My idea: F=qvBcos\theta
so the magnitude of the force is:
F=(-e)(v)(B)cos\theta

Am I correct?
What would be the relationship of the mass on this question?

This is the website: www.geocities.com/willydavidjr/emf.html

 

[nrqed]

A diagram I attached (you can view the website) shows a uniform magnetic field of magnetic flux density B[Wb/m^2] parallel to the x-axis. On the xy-plane, an electron of electric charge -e[C] and mass m[kg] departed from the x-axis in the direction of \theta[rad] with a speed of v[m/s]. Then the electron carried out a spiral movement on a cylinder whose side contains the x-axis.

This is the question: What is the magnitude of the Lorentz force acting on the electron?

My idea: F=qvBcos\theta
so the magnitude of the force is:
F=(-e)(v)(B)cos\theta

Am I correct?

Almost... it's sin(\theta) (remember, it's a cross product)

What would be the relationship of the mass on this question?

This is the website: www.geocities.com/willydavidjr/emf.html

I am not sure what you mean here...The mass does not enter the calculation of the force. If you woulsd then calculate the *acceleration* then the mass would enter, following Newton's second law.

 

[willydavidjr]

I see. I was wrong with the cosine instead of sine. I found out also that if we will calculate the radius of the circle, we will need the Newtons law of motion together with the formula of magnetic force.

This is the last question: Find he time between the electron leaving its starting point and returning to the x-axis again. Let phi denote the circular constant.

 

[marlon]

My idea: F=qvBcos\theta
so the magnitude of the force is:
F=(-e)(v)(B)cos\theta

Be sure that you keep in mind the sings here. You are using vectors ! As has been stated before the cosine should be a sine.

To get the circular motion : do you know how to acquire the equation for the orbit (the path) if you know the force ? Since, you already know what this orbit is, i suggest you look at the formula's that caracterize a circular motion. Do you know these ?

marlon

 

[willydavidjr]

I don't have any idea about the orbit but I know the magnetic force. Can you give me details on how to get this one?Thank you!

 

[nrqed]

I don't have any idea about the orbit but I know the magnetic force. Can you give me details on how to get this one?Thank you!

The motion is a sum of a motion at constant velocity in the same direction as the B field plus a circular motion at constant speed in the plane perpendicular to the B field.

So you must decompose the initial velocity in a component parallel to the B field, v_par and a component perpendicular, v_perp. For the perpendicular component just apply m {v_{perp}^2 \over r} = |q| v B sin \theta = |q| v_{perp} B since "v sin(theta)" is the component of the velocity perpendicular to the B field (at the condition that theta is the angle between the B field and the velocity vector). Using this you can find the radius of the circle. Not only that but you can find how long it takes to complete a circle which is just the period of circular motion, T = {2 \pi r \over v_{perp}} (of course, by that time the charge will have move along the x-axis so it won't get back to its initial position, it willhave moved a distance v_{par} T)

 

[willydavidjr]

Your formula for the perpendicular component m \frac {v_perp}{r} doesn't match with mass x acceleration. How is that so?

 

[nrqed]

Your formula for the perpendicular component m \frac {v_perp}{r} doesn't match with mass x acceleration. How is that so?

The equation I entered was
m { v_{perp}^2 \over r}
(as you can check by left clicking on my equation to see the Latex code). It's just that the exponent "2" is so small that it is hard to see (I should have used tex anchors instead of itex anchors).

Patrick