Electrodynamics: charge of a particle

[Gustav]

Homework Statement

We consider a part of space where there is a uniform magnetic field in the positive z-direction, $\mathbf{B}=B\hat{z}$. In addition to this magnetic field, for x < 0 there is a uniform electric field $\mathbf{E}=E\hat{y}$, in the positive y-direction. Particles with mass m are ejected in the positive x-direction from a point on the negative x-axis and then moves along the x-axis to the origin. Eventually they reach the point x = 0, y = a, z = 0. Determine the charge of the particles expressed in E, B and a!

Relevant Equations

F = q(E + v x B)

I tried solving the problem using the force formula, so what I have known is the magnetic field B and E. I also have a motion in the x-axis, that means that the velocity will be pointed at the x-axis. Inserting this in the formula I will be having something like this:
$$\mathbf{F} = q(\mathbf{E} + \mathbf{v \times B}) = q(E\hat{y} + v\hat{x}\times B\hat{z}) = q(E\hat{y} + vB \ \hat{x}\times \hat{z}) = q(E-vB)\hat{y} $$
An since the particle is moving constantly then the acceleration should be zero and the Lorentz force would be equal to zero.
$$ \mathbf{F} = q(E-vB)\hat{y} = 0 \Rightarrow E=vB$$
Now I lost myself and cannot calculate the charge. Did I do something wrong? If so what do I need to change?

 

[Gustav]

Also I thought that if the acceleration is not constant then we will be having something like this:

$$ \mathbf{F} = q(E-vB)\hat{y} = ma \Rightarrow q = \frac{ma}{E-vB} $$
But both my acceleration and velocity is unknown, so how am I going to solve this?

 

[TSny]

since the particle is moving constantly then the acceleration should be zero and the Lorentz force would be equal to zero.
$$ \mathbf{F} = q(E-vB)\hat{y} = 0 \Rightarrow E=vB$$

OK, this will be useful.

Now I lost myself and cannot calculate the charge.

Can you describe the general shape of the trajectory of the particle after it passes the origin? How does the particle get from the origin to the point (x, y, z) = (0, a, 0)?

 

[Gustav]

OK, this will be useful.Can you describe the general shape of the trajectory of the particle after it passes the origin? How does the particle get from the origin to the point (x, y, z) = (0, a, 0)?

Is a stright line from (0,0,0) to (0,a,0)?

 

[TSny]

Is a stright line from (0,0,0) to (0,a,0)?

No. Are there any forces acting on the particle when it is traveling in the region x > 0?

 

[Gustav]

No. Are there any forces acting on the particle when it is traveling in the region x > 0?

Yes, since there is no electris field in x>0
$$ \mathbf{E}=0 \Rightarrow \mathbf{F} = -qvB\hat{y} $$

 

[TSny]

Yes,
$$ \mathbf{F} = -qvB\hat{y} $$
since there is no electric field in x>0.

Right. There is no electric force in this region, but there is still a magnetic force. Your expression $\mathbf{F} = -qvB\hat{y}$ for the magnetic force is only valid if the particle is moving in the x-direction. But, the particle will be moving in the x-direction only when it first enters the region x > 0. After that, it will be deflected in a different direction by the magnetic force.

I'm sure you must have already studied the motion of a charged particle in a uniform magnetic field. Review that material.

 

[Gustav]

Right. There is no electric force in this region, but there is still a magnetic force. Your expression $\mathbf{F} = -qvB\hat{y}$ for the magnetic force is only valid if the particle is moving in the x-direction. But, the particle will be moving in the x-direction only when it first enters the region x > 0. After that, it will be deflected in a different direction by the magnetic force.

Yes, you have a point there, but since our particle will be moving in the y-direction after passing the origin then the force will be
$$ \mathbf{F} = qvB\hat{x} $$

I'm sure you must have already studied the motion of a charged particle in a uniform magnetic field. Review that material.

I have already read the material, but I am still having problem solving the question.

 

[TSny]

What is the geometric shape of the trajectory of a charged particle moving in a uniform B field if the initial velocity is perpendicular to the field?

 

[Gustav]

What is the geometric shape of the trajectory of a charged particle moving in a uniform B field if the initial velocity is perpendicular to the field?

Half-circle

 

[Gustav]

Also I can solve for $v$ given from my first expression, with other words my $v = \frac{E}{B}$

 

[TSny]

Half-circle

Yes

Also I can solve for $v$ given from my first expression, with other words my $v = \frac{E}{B}$

Yes.

 

[Gustav]

Yes

Yes.

And since the particle is moving in a "half" circular motion, then the force applied on the particle will be $\mathbf{F} = ma = \frac{mv^2}{r}$

 

[TSny]

And since the particle is moving in a "half" circular motion, then the force applied on the particle will be $\mathbf{F} = ma = \frac{mv^2}{r}$

Yes. [But to be a bit picky, the $\bf F$ on the left side should just be $F$, the magnitude of the force since $\frac{mv^2}{r}$ represents the magnitude of the force.]

 

[Gustav]

Yes. [But to be a bit picky, the $\bf F$ on the left side should just be $F$, the magnitude of the force since $\frac{mv^2}{r}$ represents the magnitude of the force.]

Oh okay, then I can do as the following.
We can set this equal to the expression in #8, and we have that
$$ F = qvB = \frac{mv^2}{r} $$
where $r$ is our radius, and we are given the diameter as the length of the line from the origin to $y = a$, so $r= \frac{a}{2}$. To solve the whole problem we have then:
$$ qvB = \frac{mv^2}{r} \Rightarrow q = \frac{mv}{rB} = \frac{m\frac{E}{B}}{\frac{a}{2}B}= \frac{2mE}{aB^2}$$

 

[TSny]

Oh okay, then I can do as the following.
We can set this equal to the expression in #8, and we have that
$$ \mathbf{F} = qvB = \frac{mv^2}{r} $$
where $r$ is our radius, and we are given the diameter as the length of the line from the origin to $y = a$, so $r= \frac{a}{2}$. To solve the whole problem we have then:
$$ qvB = \frac{mv^2}{r} \Rightarrow q = \frac{mv}{rB} = \frac{m\frac{E}{B}}{\frac{a}{2}B}= \frac{2mE}{aB^2}$$

OK. But, note that the quantities $m$, $E$, $B$, and $a$ are all positive. So, you are getting a positive answer for $q$. Can you think of a way to deduce the sign of the charge?

 

[Gustav]

OK. But, note that the quantities $m$, $E$, $B$, and $a$ are all positive. So, you are getting a positive answer for $q$. Can you think of a way to deduce whether or not the charge is positive or negative?

Can I do that using the right hand rule?

 

[TSny]

Can I do that using the right hand rule?

Yes.

 

[Gustav]

Yes.

By using the right hand rule I get that $v$ is pointing on the positive y-axis, $B$ is pointing at the positiv z-axis and $F$ is pointing at the positive x-axis. This means that the charge will also be postive?

 

[TSny]

By using the right hand rule I get that $v$ is pointing on the positive y-axis, $B$ is pointing at the positiv z-axis and $F$ is pointing at the positive x-axis. This means that the charge will also be postive?

I'm not sure I follow this. There is only one point on the semi-circular path where $\bf v$ is in the +y-direction. $\bf F$ does not point in the +x-direction at this point.

Consider the trajectory just after the particle passes the origin so that it is essentially still moving in the +x-direction. What direction must the magnetic force be at this point in order to deflect the particle in the necessary direction to eventually reach (0, a, 0)?

 

[Gustav]

I'm not sure I follow this. There is only one point on the semi-circular path where $\bf v$ is in the +y-direction. $\bf F$ does not point in the +x-direction at this point.

Consider the trajectory just after the particle passes the origin so that it is essentially still moving in the +x-direction. What direction must the magnetic force be at this point in order to deflect the particle in the necessary direction to eventually reach (0, a, 0)?

If I do it they way you adviced the magnetic force will be pointing in the negtive y-axis, and since it is pointing in the negative axis then the charge must be negative?

 

[TSny]

Yes, the charge must be negative.

 

[Gustav]

Yes, the charge must be negative.

So that means that the expression I derived in #6 is the one that I am putting equal to $\frac{mv^2}{r}$, so that I have:
$$ \mathbf{F} = -qvB\hat{y} = \frac{mv^2}{r} \Rightarrow q=-\frac{2mE}{aB^2} $$

 

[Gustav]

So that means that the expression I derived in #6 is the one that I am putting equal to $\frac{mv^2}{r}$, so that I have:
$$ \mathbf{F} = -qvB\hat{y} = \frac{mv^2}{r} \Rightarrow q=-\frac{2mE}{aB^2} $$

Since we are assuming that the velocity is pointed at the postive x-direction

 

[TSny]

Your equation $\bf{F}$ $= -qvB \hat y$ is good for the magnetic force when the particle is moving in the +x direction.

But the equation $-qvB\hat y = \frac{mv^2}{r}$ is inconsistent. On the left is a vector quantity while on the right you have a scalar quantity. You need to include a unit vector in the expression for the centripetal force to show the direction of the centripetal force just after the particle passes the origin and is moving in the +x direction.

But, you have essentially solved the problem.

 

[haruspex]

Half-circle

Well, it's a circular arc. How much of a full circle depends on the region over which the field operates and how much of the full circle fits within it.