## B-field question Lorentz's force

1. The problem statement, all variables and given/known data

3. The attempt at a solution

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

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

Mentor

## B-field question Lorentz's force

Once it completes half a circle, it's back out of the field & headed back to the left, displaced by 4 cm.
 The velocity is the left, but the force is up
 I'll give it another shot $$\frac{mv^2}{r} = qvB$$ $$v = \frac{rqB}{m}$$ $$\bar{v}t = \Delta x$$ $$\frac{rqBt}{m} = \pi r$$ $$t = \frac{\pi m}{qB}$$
 Sigh...I am alone again.
 What exactly do you mean by displaced by 4cm? Isn't it going in an arc?

Mentor
 Quote by flyingpig 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.
 Quote by flyingpig I'll give it another shot $$\frac{mv^2}{r} = qvB$$ $$v = \frac{rqB}{m}$$ $$\bar{v}t = \Delta x$$ $$\frac{rqBt}{m} = \pi r$$ $$t = \frac{\pi m}{qB}$$
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.
 Quote by flyingpig 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.
 Is my answer wrong then...? For time

Mentor
 Quote by flyingpig Is my answer wrong then...? For time
It looks like it's correct !

Find the speed, then the Kinetic Energy.
 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?
 Mentor 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?
 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
 Mentor Read my previous post about the Lorentz Force. And, YES, r is definitely 2 cm.
 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.
 Mentor There is NO other force in this problem.

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