# Ion moving through electric potential difference and magnetic field

• zenterix
zenterix
Homework Statement
A positively charged ion with charge ##q## and mass ##m##, initially at rest inside the source, is accelerated across a gap by an electric potential difference with magnitude ##\Delta v## . The ion enters a region of uniform magnetic field ##B_1## pointing into the page of the figure below and follows a semi-circular trajectory of radius ##R_1##. The ion is again accelerated across the gap, increasing its speed, by an electric potential difference that has the opposite sign but has the same magnitude ##|\Delta v|##. The ion then enters a region of uniform magnetic field ##B_2## pointing into the page of the figure below and follows a semi-circular trajectory of radius ##R_2##.
Relevant Equations
a) What is the ratio ##B_2/B_1##?

b) What is the ratio ##v_2/v_1##?
My question is about item (b).

For item (a) we have uniform circular motion in the regions with uniform magnetic field.

$$\vec{F}_{B_1}=qv\hat{\theta}_1\times B_1(-\hat{k})=-qv_1B_1\hat{r}_1=-mR_1\theta'^2\hat{r}_1\tag{1}$$

$$B_1=\frac{mv_1}{R_1q}\tag{2}$$

A similar calculation for the bottom portion of the trajectory gives

$$B_2=\frac{mv_2}{R_2q}\tag{3}$$

Thus,

$$\frac{B_2}{B_1}=\frac{R_1v_2}{R_2v_1}\tag{4}$$

As for item (b), I don't understand the following.

The problem statement says that the ion is accelerated in both of the gaps. It also says that the electric potential difference in the gap on the left is the negative of the electric potential difference of the gap on the right.

How then can a positively charged ion accelerate across both gaps?

Let's assume that the potential at the "source" is higher than on the top part of the gap above it.

$$q\Delta v=-W=-\Delta K_e=-\frac{mv_1^2}{2}\tag{5}$$

$$v_1=\sqrt{-\frac{2q}\Delta V}{m}}\tag{6}$$

Note that ##\Delta v<0##.

Now consider the gap on the left side.

The speed of the ion is the same ##v_1## when it enters the gap, but now it is traversing from a lower to a higher potential as a positive charge. Shouldn't it slow down (to zero speed)?

Note that the problem states that the speed increases during both acceleration stages and the potential difference in the drawing is a magnitude. What does that indicate to you?

MatinSAR
zenterix said:
$$\frac{B_2}{B_1}=\frac{R_1v_2}{R_2v_1}\tag{4}$$
I don't think the velocities should figure in the answer. Can you see how to get rid of those?
zenterix said:
It also says that the electric potential difference in the gap on the left is the negative of the electric potential difference of the gap on the right.
It must mean the p.d. is in the opposite direction relative to the diagram, so in the same direction relative to the path.

MatinSAR
zenterix said:
$$v_1=\sqrt{-\frac{2q\Delta V} {m}}\tag{6}$$
Note that ##\Delta v<0##.
Note that the use of ##v## for both speed and voltage is error-prone ! So is the use of ##\Delta v## and ##\Delta V## for one and the same variable

zenterix said:
the speed of the ion is the same v1 when it enters the gap, but now it is traversing from a lower to a higher potential as a positive charge. Shouldn't it slow down (to zero speed)?
That's not the intention:
zenterix said:
The ion is again accelerated across the gap, increasing its speed, by an electric potential difference that has the opposite sign but has the same magnitude ##|\Delta v|##.
but I grant you the wording is confusing. I think they have a ##\Delta V \equiv V_{top}-V_{bot} ## definition.

The cyclotron principle is explained here

##\ ##

haruspex said:
I don't think the velocities should figure in the answer. Can you see how to get rid of those?
Why don't you think the velocities should be there?

This problem has an automatic grading system. The expression for ##\frac{B_2}{B_1}## checks out.

There is an item (c) to this problem that I did not show. It is to suppose that ##R_2=2R_1## and compute what ##\frac{B_2}{B_1}## is.

Since ##\frac{B_2}{B_1}## has a ##\frac{v_2}{v_1}## term in it, which in item (b) comes out to be ##\sqrt{2}##, we can sub this in to get ##\frac{B_2}{B_1}=\frac{\sqrt{2}}{2}##.

BvU said:
Note that the use of v for both speed and voltage is error-prone ! So is the use of Δv and ΔV for one and the same variable
I am using ##v## for speed and ##\Delta V## for potential difference. The use of ##\Delta v## in the OP is a typo.

zenterix said:
Since ##\frac{B_2}{B_1}## has a ##\frac{v_2}{v_1}## term in it, which in item (b) comes out to be ##\sqrt{2}##, we can sub this in to get ##\frac{B_2}{B_1}=\frac{\sqrt{2}}{2}##.
That's why I thought the velocities should not appear in the first answer.

MatinSAR and zenterix

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