Motion of a Charged Particle in a Constant Magnetic Field

[etcicco]

Homework Statement

A charged particle of mass m = 6.4X10E-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 3T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.84 m, 0) and leaves the region at (x,y) = 0, 0.84 m a time t = 694 μs after it entered the region.

1) With what speed v did the particle enter the region containing the magnetic field?

2) What is Fx, the x-component of the force on the particle at a time t1 = 231.3 μs after it entered the region containing the magnetic field.

3) What is Fy, the y-component of the force on the particle at a time t1 = 231.3 μs after it entered the region containing the magnetic field.

4) What is q, the charge of the particle? Be sure to include the correct sign.

5) If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same?

Homework Equations

F = ma => a = (v^2)/r
F = qvB
qvB = m(v^2)/r

The Attempt at a Solution

Question 1 Solved: v = 1901.25 m/s

Question 2: Found the distance at which it traveled, by multiplying time by the velocity. And from there i used arc length divided by radius to get theta. So, θ = s/r. Then from there I used F = m*(v^2)/r * cosθ to try and solve for the x direction.

Question 3: All I did was used F = m*(v^2)/r * θ, This answer came out to be right, how ever I do not understand why it is right. So can some one please explain to me.

Question 4: I got 48.2857, how ever I do not know if this is right, and I have no way to check.

Question 5: I said increase B by a factor of 2, how ever I do not know if this is right, and again I have no way to check it.

 

[BvU]

1) OK. You have R in two digits, t in 3. So v in 6 digits is somewhat exaggerated. 1900 or 1901 m/s would be adequate.

2) did you notice that this is 1/3 the time it took the particle to do 1/4 circle ? So $\theta = \pi / 3$ would be a little more direct ?

3) what is the difference with 2 ?

I notice you write all your relevant equations a simple products. Which is Ok if you do keep in mind that in fact they are vector equations:

F = ma => a = (v^2)/r is $\quad \vec F = m \vec a \quad \buildrel ? \over \rightarrow \quad |\vec a| = {\vec v \cdot \vec v \over |\vec r|}$ (I wrote a ?, because I don't see the direct consequence )
F = qvB is $\quad \vec F = q \ \vec v \times \vec B \quad$ with x the outer product

qvB = m(v^2)/r $\quad \quad$ see above

And now the interpretation: The Lorentz force is at all times perpendicular to $\vec v$ and $\vec B$, so it does not change $| \vec v |$. With both $| \vec v |$ and $| \vec B |$ constant, the trajectory is an arc with a constant radius, as per your eqn.

At all points of the trajectory the force is perpendicular to the radius vector. (So: once you have $\theta$, you have $(\vec F_x, \vec F_y )\quad$ ).

Question 4: you have m, v, B, r (yes, $| \vec v |$ etc.) and you have an equation. Why do you say you have no way to check ?

5. same as 4.

 

[BvU]

And:

F = m*(v^2)/r * θ, This answer came out to be right, ...

is a bit strange. No sines involved, no sign involved either ?
How do you know 2 and 3 are right ? I would expect negative numbers, but your expressions only allow positive outcomes ?

 

[etcicco]

1) OK. You have R in two digits, t in 3. So v in 6 digits is somewhat exaggerated. 1900 or 1901 m/s would be adequate.

2) did you notice that this is 1/3 the time it took the particle to do 1/4 circle ? So $\theta = \pi / 3$ would be a little more direct ?

3) what is the difference with 2 ?

I notice you write all your relevant equations a simple products. Which is Ok if you do keep in mind that in fact they are vector equations:

F = ma => a = (v^2)/r is $\quad \vec F = m \vec a \quad \buildrel ? \over \rightarrow \quad |\vec a| = {\vec v \cdot \vec v \over |\vec r|}$ (I wrote a ?, because I don't see the direct consequence )
F = qvB is $\quad \vec F = q \ \vec v \times \vec B \quad$ with x the outer product

qvB = m(v^2)/r $\quad \quad$ see above

And now the interpretation: The Lorentz force is at all times perpendicular to $\vec v$ and $\vec B$, so it does not change $| \vec v |$. With both $| \vec v |$ and $| \vec B |$ constant, the trajectory is an arc with a constant radius, as per your eqn.

At all points of the trajectory the force is perpendicular to the radius vector. (So: once you have $\theta$, you have $(\vec F_x, \vec F_y )\quad$ ).

Question 4: you have m, v, B, r (yes, $| \vec v |$ etc.) and you have an equation. Why do you say you have no way to check ?

5. same as 4.

Okay, this is putting me on the right track. Thanks for the fast response. I am going to start with the first one I need to do, question 2. I did not notice that it is 1/3 the time. What did you mean θ/3? I am confused on where I would be using the theta/3? would I multiply the force by cos(θ/3)?

 

[BvU]

Oh, and welcome to PF! Your use of the template is exemplary, so I could get straight to the point. Hope it helps...

 

[BvU]

Did I mistype $\theta / 3$ somewhere?

I did mistype one third of 90 degrees. It should be $\pi / 6$. But you already fixed that I suppose.

 

[etcicco]

Thank you! And no. This whole lesson of magnetism is confusing me. I broke my hand and had to miss a few days of class. So I am jumping straight into it on my own

 

[BvU]

Well, typing with a cast on your hand isn't the easiest thing I can think of, but you seem to do OK.
Any questions? PF, or google around (Lorentz force, or: charged particle in magnetic field)
But I take it you also have a textbook. So happy circling!