Fundamental Forces: finding velocity using different fields and forces

[JoeyBob]

Homework Statement

See attached

Relevant Equations

F=qE, F=qv cross product B

So I know the acceleration is 0, so the net force is 0.

QE=1.6E-3*9.9E3 k hat = 15.84 k hat (thats one force)

qv x B = q(v i hat + v j hat) x (0.51 i hat + 0.10 j hat)

=q(0.10 v k hat- 0.51 v k hat)
=q(-0.41 v k hat)
=-0.000656 v k hat

Now solving for velocity,

0.000656v=15.84, v=24146.34146

The question says I need to find the ratio, and I know vy is .41 times higher, so i multiply by this to get 9890.

This answer is wrong, it should be -3665.31.

 

[haruspex]

v i hat + v j hat

I don't understand what you mean by this. Elsewhere, you interpret v as the scalar speed, but above you seem to mean $v_i\hat i+v_j\hat j$.
Btw, there is a neater way than the given hint.
Given equations $\vec X\times\vec Y=\vec Z$ and $\vec X.\vec Y=A$ you can take the triple vector product $\vec Y\times\vec Z=\vec Y\times(\vec X\times\vec Y)=(\vec Y.\vec Y)\vec X-(\vec Y.\vec X)\vec Y=Y^2\vec X-A\vec Y$.
Whence $\vec X= \frac{\vec Y\times\vec Z+A\vec Y}{Y^2}$.

 

[JoeyBob]

I don't understand what you mean by this. Elsewhere, you interpret v as the scalar speed, but above you seem to mean $v_i\hat i+v_j\hat j$.
Btw, there is a neater way than the given hint.
Given equations $\vec X\times\vec Y=\vec Z$ and $\vec X.\vec Y=A$ you can take the triple vector product $\vec Y\times\vec Z=\vec Y\times(\vec X\times\vec Y)=(\vec Y.\vec Y)\vec X-(\vec Y.\vec X)\vec Y=Y^2\vec X-A\vec Y$.
Whence $\vec X= \frac{\vec Y\times\vec Z+A\vec Y}{Y^2}$.

How are you suppose to use the hint?

 

[kuruman]

Perhaps it would be conceptually easier for you to write two equations, one saying that the net force on the particle is zero and another saying that the velocity is perpendicular to the magnetic field. You will have two equations and two unknowns, $v_x$ and $v_y$.

Don't forget that the net force is the sum of an electric force and a magnetic force.

 

[JoeyBob]

Perhaps it would be conceptually easier for you to write two equations, one saying that the net force on the particle is zero and another saying that the velocity is perpendicular to the magnetic field. You will have two equations and two unknowns, $v_x$ and $v_y$.

Don't forget that the net force is the sum of an electric force and a magnetic force.

So one force would be F=qE. I think its the magnetic field force that is confusing me.

Would it just be F=q($v_x$+$v_y$) x B?

Then these forces added together are 0?

Thats what I did when I tried to solve it, but I got stuck because I have two unknown variables now (the two different velocity vectors).

 

[pai535]

qv x B = q(v i hat + v j hat) x (0.51 i hat + 0.10 j hat)

=q(0.10 v k hat- 0.51 v k hat)
=q(-0.41 v k hat)
=-0.000656 v k hat

You made a mistake in this calculation of cross product. Maybe you can write down the determinant explicitly and try to figure out what are the components.

Thats what I did when I tried to solve it, but I got stuck because I have two unknown variables now (the two different velocity vectors).

There is another hint in your question that you didn't use - the velocity is perpendicular to magnetic field. As mentioned above, you will have two variables and two equations.

 

[kuruman]

Would it just be F=q($v_x$+$v_y$) x B?

It would not. Please (re)view how to write the cross product in terms of its components here to see what it would be.

Thats what I did when I tried to solve it, but I got stuck because I have two unknown variables now (the two different velocity vectors).

I know one velocity vector, the one associated with the particle mentioned in the problem which is $\vec v=v_x~\hat x+v_y~\hat y.$ What velocity vector is the other one and what entity has it?

 

[haruspex]

How are you suppose to use the hint?

My vector triple product hint or the one they gave?

 

[JoeyBob]

You made a mistake in this calculation of cross product. Maybe you can write down the determinant explicitly and try to figure out what are the components.
There is another hint in your question that you didn't use - the velocity is perpendicular to magnetic field. As mentioned above, you will have two variables and two equations.

What equation is that?

 

[kuruman]

What equation is that?

$\vec B\cdot \vec v=0.$

 

[JoeyBob]

$\vec B\cdot \vec v=0.$

Wouldn't that just mean the velocity is 0 then?

 

[kuruman]

No. It's another way of saying that the two vectors are perpendicular. I suggest tht you review dot products and cross products.

 

[JoeyBob]

No. It's another way of saying that the two vectors are perpendicular. I suggest tht you review dot products and cross product.

So then the velocity is either -1.96 i hat + 10 j hat or 1.96 i hat - 10 j hat if its perpendicular. But this doesn't really help me.

 

[kuruman]

So then the velocity is either -1.96 i hat + 10 j hat or 1.96 i hat - 10 j hat if its perpendicular. But this doesn't really help me.

How did you arrive at these numbers? Are these the only choices that make the magnetic field and velocity perpendicular? Remember that the particle has to go in a straight line. As I mentioned before, the net force on the particle must be zero because we are told that it is moving in a straight line. See post #4.

 

[JoeyBob]

How did you arrive at these numbers?

Because with the dot product they give 1-1, which is 0. I don't see why it wouldn't be a straight line, it would go in the direction of the vectors.

 

[kuruman]

Which direction is this and which vectors are you talking about? There is an electric field along the z-axis and the magnetic field in the xy plane. The particle's velocity is perpendicular to both the electric and magnetic field. Therefore it is not movinfg along either field.

Having the velocity perpendicular to the field is not enough to ensure that the particle is moving in a straight line. That's because the magnetic force on the particle is velocity-dependent. The velocity has to be just right so that the sum of the electric and magnetic forces on the particle is zero.

 

[JoeyBob]

Which direction is this and which vectors are you talking about? There is an electric field along the z-axis and the magnetic field in the xy plane. The particle's velocity is perpendicular to both the electric and magnetic field. Therefore it is not movinfg along either field.

Having the velocity perpendicular to the field is not enough to ensure that the particle is moving in a straight line. That's because the magnetic force on the particle is velocity-dependent. The velocity has to be just right so that the sum of the electric and magnetic forces on the particle is zero.

I know B=0.51 i hat + 0.10 j hat. I need to find the ratio of vx to vy by making the dot product =0.

(0.51 i hat + 0.10 j hat) * (1.96 i hat -10 j hat) =0. (0.51 i hat + 0.10 j hat) * (-1.96 i hat + 10 j hat) is also 0. I assume this means that the vx has a ratio of 0.196 to vy.

 

[kuruman]

Sure, but that is not enough to determine vx and vy separately. It is also true that
(0.51 i hat + 0.10 j hat) * (2*1.96 i hat -2*10 j hat) = 0
(0.51 i hat + 0.10 j hat) * (23.4*1.96 i hat -23.4*10 j hat) = 0
(0.51 i hat + 0.10 j hat) * (1065*1.96 i hat -1065*10 j hat) = 0
...
And so on. You get the picture? Like I said, of all these infinitely many choices of velocities perpendicular to the magnetic field, only one is correct and will result in the particle following a straight line. The equation saying that the net force is zero will give that correct choice.

 

[JoeyBob]

Sure, but that is not enough to determine vx and vy separately. It is also true that
(0.51 i hat + 0.10 j hat) * (2*1.96 i hat -2*10 j hat) = 0
(0.51 i hat + 0.10 j hat) * (23.4*1.96 i hat -23.4*10 j hat) = 0
(0.51 i hat + 0.10 j hat) * (1065*1.96 i hat -1065*10 j hat) = 0
...
And so on. You get the picture? Like I said, of all these infinitely many choices of velocities perpendicular to the magnetic field, only one is correct and will result in the particle following a straight line. The equation saying that the net force is zero will give that correct choice.

But the ratio is always the same, no?

I understand that 0=qE+qv x B, but idk how to use the ratios.

 

[kuruman]

Write the equation $q\vec E+q\vec v\times \vec B=0$ correctly in terms of $v_x$ and $v_y$. I gave you a link in post #7 if you don't know how to do that. That is the second equation, the first one being $B_x v_x+B_y v_y=0.$

 

[pai535]

But the ratio is always the same, no?

I understand that 0=qE+qv x B, but idk how to use the ratios.

Using the ratio, you can express $v_y$ in terms of $v_x$ (also $v_x$ in terms of $v_y$). Then you reduce the problem to a problem with only one unknown variable.

 

[JoeyBob]

Using the ratio, you can express $v_y$ in terms of $v_x$ (also $v_x$ in terms of $v_y$). Then you reduce the problem to a problem with only one unknown variable.

okay here's what I did.

Knowing B x (vx + vy)=0, I found the ratio of vx to vy was -0.196

Now I also know 0=qE+qv x B

-E=v x B

-E=(vx i hat -vx/0.196 j hat) x (0.51 i hat + 0.10 j hat)

-E=0.10vx k hat+2.602vx k hat

-E/(0.1+2.602)=vx=-3664

This is one less than the right answer so I am assuming i made some rounding error.

Thanks for the help everyone! Took me a while but I got it...

 

[kuruman]

B x (vx + vy)=0,

For future reference, use the symbol "x" to denote cross product only. Here the context is dot product, so use a dot. Also, use unit vectors where applicable. (vx + vy) looks like you are adding the two components of a vector as scalars, not as vectors n orthogonal directions. This is important if you seek help and need to be understood because although you know what you mean, the reader may not.

I would strongly recommend that you invest some time to learn how to use LaTeX to write equations. It's a skill that will serve you well if you follow a technical career. PF provides a good LaTeX starter guide if you click "LaTeX Guide" on this screen, bottom left, just above "Attach files".

 

[JoeyBob]

For future reference, use the symbol "x" to denote cross product only. Here the context is dot product, so use a dot. Also, use unit vectors where applicable. (vx + vy) looks like you are adding the two components of a vector as scalars, not as vectors n orthogonal directions. This is important if you seek help and need to be understood because although you know what you mean, the reader may not.

I would strongly recommend that you invest some time to learn how to use LaTeX to write equations. It's a skill that will serve you well if you follow a technical career. PF provides a good LaTeX starter guide if you click "LaTeX Guide" on this screen, bottom left, just above "Attach files".

Im doing a chemistry focused degree, so idk how beneficial learning LaTeX would be in the furute. Have to take first year physics then I am done with it.

Personally physics isn't that bad, but its not something I would want to pursue a career in.

 

[pai535]

Im doing a chemistry focused degree, so idk how beneficial learning LaTeX would be in the furute. Have to take first year physics then I am done with it.

Personally physics isn't that bad, but its not something I would want to pursue a career in.

Actually LaTeX is everywhere in science, not just physics. If you want to write and publish scientific articles (and even books), you'd better learn some LaTeX. Probably there will also be formulas and calculations in your future study (although I don't know much about chemistry). Using LaTeX can make them much easier to read.

 

[haruspex]

Took me a while but I

Here's my method:
$\vec E+\vec v\times\vec B=0$
$\vec v.\vec B=0$
Taking the cross product of the first equation with $\vec B$:
$0=\vec B\times\vec E+\vec B\times(\vec v\times\vec B)=\vec B\times\vec E+(\vec B.\vec B)\vec v-(\vec B.\vec v)\vec B=\vec B\times\vec E+(\vec B.\vec B)\vec v$
Whence $\vec v=-\frac{\vec B\times\vec E}{\vec B.\vec B}$

 

[kuruman]

Here's my method:
$\vec E+\vec v\times\vec B=0$
$\vec v.\vec B=0$
Taking the cross product of the first equation with $\vec B$:
$0=\vec B\times\vec E+\vec B\times(\vec v\times\vec B)=\vec B\times\vec E+(\vec B.\vec B)\vec v-(\vec B.\vec v)\vec B=\vec B\times\vec E+(\vec B.\vec B)\vec v$
Whence $\vec v=-\frac{\vec B\times\vec E}{\vec B.\vec B}$

Alternatively, the set up can be recognized as a velocity selector in which case the speed is equal to the ratio of the field magnitudes and the direction perpendicular to both crossed fields such that the electric and magnetic forces are antiparallel. Then, $$v=\frac{E}{B}~;~~\hat v=\frac{\vec E\times \vec B}{EB}~~\Rightarrow~\vec v=v~\hat v=\frac{\vec E\times \vec B}{B^2}.$$Note that the three participating unit vectors form a right-handed coordinate system. The result is independent of Cartesian axes and of the sign of the moving charge.

 

[JoeyBob]

Alternatively, the set up can be recognized as a velocity selector in which case the speed is equal to the ratio of the field magnitudes and the direction perpendicular to both crossed fields such that the electric and magnetic forces are antiparallel. Then, $$v=\frac{E}{B}~;~~\hat v=\frac{\vec E\times \vec B}{EB}~~\Rightarrow~\vec v=v~\hat v=\frac{\vec E\times \vec B}{B^2}.$$Note that the three participating unit vectors form a right-handed coordinate system. The result is independent of Cartesian axes and of the sign of the moving charge.

Above my level I think.

 

[pai535]

Above my level I think.

Just think of an apparatus that if you send in a beam of charged particles, only the ones with a specific speed can travel through. Then you can sort out all the particles with the speed you want (by adjusting E and M fields).