B-field question Lorentz's force

[flyingpig]

Homework Statement

http://img834.imageshack.us/img834/5598/46421047.th.png

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The Attempt at a Solution

I think this problem is deeper than it looks. Here is my FBD

http://img291.imageshack.us/img291/69/31024385.th.png

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So I am assuming at it actually "stops" and "drops" when the magnetic force is exactly 180 degrees with Fg and then it falls under constant velocity?

In other words

$$qv_{0}B = mg$$

$$v_0 = \frac{mg}{qB}$$

Then

$$y(t) = \frac{mgt}{qB} + y(0)$$

Now my problem is, what is y(0)?

 

[SammyS]

I suspect you are to ignore gravity. At any rate the effect of gravity will be minuscule compared to the Lorentz force.

The electron will undergo uniform circular motion once it enters the region with the B field. It's path will be a semi-circle, after which it exits the field.

What is the direction of the Lorentz force on the electron?

 

[flyingpig]

It's always inwards, if there is no gravity then won't it go into a circle forever?

 

[SammyS]

Once it completes half a circle, it's back out of the field & headed back to the left, displaced by 4 cm.

 

[flyingpig]

The velocity is the left, but the force is up

 

[flyingpig]

I'll give it another shot

$$\frac{mv^2}{r} = qvB$$

$$v = \frac{rqB}{m}$$

$$\bar{v}t = \Delta x$$

$$\frac{rqBt}{m} = \pi r$$

$$t = \frac{\pi m}{qB}$$

 

[flyingpig]

Sigh...I am alone again.

 

[flyingpig]

What exactly do you mean by displaced by 4cm? Isn't it going in an arc?

 

[SammyS]

The velocity is the left, but the force is up

$$\vec{v}$$ is to the right, so $$-e\vec{v}$$ is to the left, so, yes, the force is upward, on the page, initially, where q = -e.

I'll give it another shot

$$\frac{mv^2}{r} = qvB$$

$$v = \frac{rqB}{m}$$

$$\bar{v}t = \Delta x$$

$$\frac{rqBt}{m} = \pi r$$

$$t = \frac{\pi m}{qB}$$

The radius is given in the statement of the problem -- although not directly. So, you should be able to find the velocity, and thus the Kinetic Energy.

What exactly do you mean by displaced by 4cm? Isn't it going in an arc?

After the electron leaves the region with the B-field, it again travels in a straight line. This line is 4 cm from the electron's initial line of flight.

 

[flyingpig]

Is my answer wrong then...? For time

 

[SammyS]

Is my answer wrong then...? For time

It looks like it's correct !

Find the speed, then the Kinetic Energy.

 

[flyingpig]

Sammy, I am still confused about the KE part, is it asking when it comes out? Or the change in KE during the whole process? If it is the whole trip, how do I find the tangential force?

 

[SammyS]

Lorentz Force is perpendicular to the direction of motion (cross product) at all times. Therefore, the Lorentz Force does NO work on the electron, so KE is constant.

Io find KE, you need to find v. What is r for the circular portion of the motion?

 

[flyingpig]

I feel like you are trying to lead me to saying it is 2cm...but it isn't. When I mean tangential force, I don't mean the centripetal force I mean the one that's parallel to the tangential velocity

 

[SammyS]

Read my previous post about the Lorentz Force.

And, YES, r is definitely 2 cm.

 

[flyingpig]

No the Lorentz Force = centripetal force, which not the force I am talking about.

How do I find the force that is parallel to the tangential velocity si what I am asking.

 

[SammyS]

There is NO other force in this problem.

 

[flyingpig]

How do I find the work done = KE then?

$$ma_t \cdot \pi r = \Delta KE$$

 

[SammyS]

The Lorentz Force is the only force in this problem. The Lorentz Force does NO work on the electron, as has already been explained.

Therefore, ΔKE =0.

 

[flyingpig]

http://img190.imageshack.us/img190/7383/jok.th.png

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[SammyS]

http://img190.imageshack.us/img190/7383/jok.th.png

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What part of "only" don't you understand?

The Lorentz Force is the only force in this problem.

The force you seem to be looking for doesn't exist.

 

[flyingpig]

Then how can I find the change KE when it is 0? I don't understand

 

[clementc]

It's meant to just be the KE I think - there's no change in KE
There's only centripetal force in this problem - the "tangential force" you're talking about doesn't exist =\

 

[SammyS]

You can lead a horse to water, but you can't make him drink>

Review my posts. You have all the information you need.

Change in KE is zero, KE ≠ 0, but it doesn't change!

 

[flyingpig]

I don't understand how the 2cm can help me get to the answer. If the change in KE is 0 then

$$\frac{1}{2}mv_{0}^2 = \frac{1}{2}mv^2$$

 

[SammyS]

I'll give it another shot

$$\frac{mv^2}{r} = qvB$$

$$v = \frac{rqB}{m}$$

$$\bar{v}t = \Delta x$$

$$\frac{rqBt}{m} = \pi r$$

$$t = \frac{\pi m}{qB}$$

I don't understand how the 2cm can help me get to the answer. If the change in KE is 0 then

$$\frac{1}{2}mv_{0}^2 = \frac{1}{2}mv^2$$

One of the equations in post #6 will give you the speed.

You know r, q, B, & m. Find v.

 

[flyingpig]

I was about to say that 2.00cm / t = v...

$$F = qvB = m\frac{v^2}{r}$$

$$qB = m\frac{v}{r}$$

$$\frac{qBr}{m} = v$$

 

[flyingpig]

$$KE = \frac{1}{2}(qBr)^2$$

Now the question is, is r the ARCLENGTH or radius. How did you know it was radius Sammy?

 

[SammyS]

$$KE = \frac{1}{2}(qBr)^2$$

You're missing an m.

Now the question is, is r the ARCLENGTH or radius. How did you know it was radius Sammy?

The equation: $$F = qvB = m\frac{v^2}{r}$$ tells us that the Lorentz force is providing the centripetal force. Since m is mass & v is the speed, r must be the radius of the resulting semi-circle.

Before you ask me, I'm asking you:

flyingpig, why is it a semi-circle rather than a full circle?​

 

[flyingpig]

You're missing an m.

The equation: $$F = qvB = m\frac{v^2}{r}$$ tells us that the Lorentz force is providing the centripetal force. Since m is mass & v is the speed, r must be the radius of the resulting semi-circle.

Before you ask me, I'm asking you:

flyingpig, why is it a semi-circle rather than a full circle?​

Because there is no B-field on the left hand side...it says on the picture

 

[flyingpig]

Ah Yes, the m lol

 

[SammyS]

Are we squared away?

You should be able to give a numerical answer.

 

[flyingpig]

a) 1.79 x 10^-8s

b) 5.63 x 10^-18J

Unless I made a mistake typing in my numbers.Sammy, I also noticed that you could do this

The arc length is $$\pi (2.00 \times 10^-2)$$

The speed is then

$$\frac{\pi (2.00 \times 10^-2)}{1.79 \times 10^-8}$$

Then I also get the same KE!

 

[flyingpig]

Also, Sammy, when I calculated the velocity I used the average velocity formula and I forgot to divide it by 2. I should have used the change in velocity. My question, how come they don't give the same answer?

 

[SammyS]

The speed doesn't change.