Using Lorentz Force to derive V = E/B

[phantomvommand]

Homework Statement

I am trying to use the Lorentz Force formula to derive v = E/B. What is wrong with my working?

Relevant Equations

$F = q(\mathbf E + \mathbf v \times \mathbf B)$

Suppose the E-field is $-E_y\hat y$, and B-field is $B\hat z$. Mass is $m$.
z
|
|_____x
/
y
$m(\ddot x \hat x + \ddot y \hat y) = q(-E_y \hat y + (v_x \hat x + v_y \hat y) \times B \hat z)$
By grouping terms with $\hat x$ and $\hat y$ together,
$m\ddot x = -qv_yB$
$m\ddot y = q(v_xB - E_y)$

Set $\ddot x = 0, \ddot y = 0$,
$-v_yB = v_xB - E_y$.

Why don't I get $v = \frac E B$?

 

[Astronuc]

What is significant about the velocity in relation to the applied electric field and magnetic field?

 

[TSny]

Suppose the E-field is $-E_y\hat y$, and B-field is $B\hat z$. Mass is $m$.

It might be better to write the electric field as $\mathbf E = -E \, \hat y$, rather than $\mathbf E = -E_y \, \hat y$. The symbol $E_y$ usually denotes the entire coefficient of $\hat y$.

$m\ddot x = -qv_yB$
$m\ddot y = q(v_xB - E_y)$

Check the signs for the two terms that are proportional to B.

Set $\ddot x = 0, \ddot y = 0$,
$-v_yB = v_xB - E_y$.

Write two separate equations. One for $\ddot x = 0$ and one for $\ddot y = 0$. By combining these into one equation, you lost some important information.

In the statement of the problem, I assume you want the force on q to be zero.

 

[phantomvommand]

It might be better to write the electric field as $\mathbf E = -E \, \hat y$, rather than $\mathbf E = -E_y \, \hat y$. The symbol $E_y$ usually denotes the entire coefficient of $\hat y$.Check the signs for the two terms that are proportional to B.
Write two separate equations. One for $\ddot x = 0$ and one for $\ddot y = 0$. By combining these into one equation, you lost some important information.

In the statement of the problem, I assume you want the force on q to be zero.

$0 = qv_yB$
Does the above imply $v_y = 0$?
$0 = qv_xB - qE$
The above implies $v_x = \frac E B$.
Indeed, I think I am supposed to get $v_x = \frac E B$ instead of $v = \frac E B$.
I have used $E$ instead of $E_y$.
This should resolve it, thanks so much!

 

[TSny]

$0 = qv_yB$
Does the above imply $v_y = 0$

Yes. (We may assume that neither q nor B is zero.)

$0 = qv_xB - qE$

Check the sign of the first term on the right side of $0 = qv_xB - qE$

The above implies $v_x = \frac E B$.

Since E and B are the magnitudes of the fields, they are positive numbers. So, you are getting that $v_x$ is positive. By drawing a sketch and using the right-hand rule for the cross product for the magnetic force, do you expect a positive value of $v_x$?

Is it possible to deduce $v_z$?

 

[TSny]

I just noticed why there is some confusion with signs.

z
|
|_____x
/
y

You are using a left-handed coordinate system. I'm in the habit of assuming a right-handed system.

So, I think your result for $v_x$ is correct for your coordinate system.

 

[Steve4Physics]

The problem-statement is incomplete. Presumably this is a simple 'velocity selector' (Wien filter) with $\vec v, \vec E$ and $\vec B$ mutually perpendicular.

In this case you require $\vec v$ to be unchanged by the selector. This means $\vec F$ and $\vec a$ must both be zero. So to make the working neater you really only need to solve $\vec E + \vec v \times \vec B = 0$.

Without loss of generality, you can choose x, y and z axes such that $\vec v, \vec E$ and $\vec B$ each lie along an axis. The problem is then simple. (Though I may have the wrong end of the stick and you may be deliberately doing it in terms of components as an exercise.)

 

[phantomvommand]

I just noticed why there is some confusion with signs.
You are using a left-handed coordinate system. I'm in the habit of assuming a right-handed system.

So, I think your result for $v_x$ is correct for your coordinate system.

I don’t think it’s possible to deduce $v_z$ , is it?

 

[TSny]

I don’t think it’s possible to deduce $v_z$ , is it?

That's right. The force felt by q is independent of $v_z$.