Which Direction Does the Force on the Electrons Act in a TV Tube?

[mshiddensecret]

Homework Statement

The electrons in the beam of a television tube have an energy of 15 keV. The tube is oriented so that the electrons move horizontally from south to north. The vertical component of the Earth's magnetic field at the location of the television has a magnitude of 57.0 μT and is pointing down.

a) In which direction does the force on the electrons act (enter N for north, S for South, E for East, or W for West)?
Neglect a possible horizontal component of the magnetic field.

b) What is the magnitude of the acceleration due to the vertical component of the Earth's magnetic field of an electron in the beam?

c) If the inclination of the Earth's magnetic field near the TV is 66deg, calculate the magnitude of the force on the electrons due to the horizontal component of the Earth's magnetic field.

Homework Equations

B = qv x B

The Attempt at a Solution

a) according to the right hand rule, it should be W. But apparently i am wrong.

b) I got the answer which is 7.26 x 10^14 m/s^2

and also the F = 6.62 x 10^-16 N

c) it should be 6.62x10^-16 x sin 66 right? I didn't get it right.

 

[Simon Bridge]

(a) how are you applying the right-hand rule?
Did you account for electrons being negatively charged?

(b) are you saying you got the right answer?

(c) so what did you get and how did you get it?

 

[mshiddensecret]

A) I found out. It is east. B) I got the correct answer

C) I multiply the force which I got before dividing it by the mass to get the acceleration. And multiply by sin 66 which is incorrect. I am lost

 

[cheah10]

For "a)", how do you know you are wrong?

For "b)", can you tell us what working did you do to reach the answer for acceleration and force?

 

[cheah10]

ignore my previous post

 

[Simon Bridge]

A) I found out. It is east.

... yes, but how did you apply the right-hand rule so that you got "west" ... the RHR should give you east.
Basically it sounds like you did the right-hand rule correctly for the cross product, but neglected that the electron is negative.
The equation is $\vec F=q \vec v \times \vec B$ and in your case, $q=-e$. That minus sign reverses the direction of the resultant.

B) I got the correct answer

Well done. I take it you used $F_B=qvB$?

C) I multiply the force which I got before dividing it by the mass to get the acceleration. And multiply by sin 66 which is incorrect. I am lost

...
So you used: $F_C = qvB\sin\theta = F_B \sin\theta$ where $F_B$ is the force from part (B)?
If this answer is wrong then you probably have the wrong value for the angle between $\vec v$ and $\vec B.$.
If the answer is being marked by a computer, then you also need to check for rounding errors and the correct sig fig. Stuff like that.

 

[mshiddensecret]

... yes, but how did you apply the right-hand rule so that you got "west" ... the RHR should give you east.
Basically it sounds like you did the right-hand rule correctly for the cross product, but neglected that the electron is negative.
The equation is $\vec F=q \vec v \times \vec B$ and in your case, $q=-e$. That minus sign reverses the direction of the resultant.

Well done. I take it you used $F_B=qvB$?

...
So you used: $F_C = qvB\sin\theta = F_B \sin\theta$ where $F_B$ is the force from part (B)?
If this answer is wrong then you probably have the wrong value for the angle between $\vec v$ and $\vec B.$.
If the answer is being marked by a computer, then you also need to check for rounding errors and the correct sig fig. Stuff like that.

What do you mean by the wrong value for the angle? 66?

 

[mshiddensecret]

6.05e-16 N was my answer for C.

 

[Simon Bridge]

What do you mean by the wrong value for the angle? 66?

The problem statement says that the inclination of the field is 66deg.
It does not say that this is the same as the angle between the velocity and field vectors - your result would imply that it isn't.
So what is it an angle to?

It's always possible that the computer answer is wrong - it happens.
It could also be a mistake in your arithmetic ... I don't check arithmetic.

6.05e-16 N was my answer for C.

Bare numbers don't mean anything. What counts is how you got them - by what reasoning.

 

[mshiddensecret]

I have no idea at this point.

 

[Simon Bridge]

Well if 66deg is not the angle of the field to the velocity, what could it be to?
Have you tried googling for "inclination of Earth's magnetic field" to find out what most people mean by that?

Have you checked your arithmetic?
Have you checked for rounding errors?
Have you checked you have the right decimal places?

"no idea" is not acceptable - you have the means to get ideas: use them.

 

[mshiddensecret]

24?

 

[Simon Bridge]

24?

Bare numbers have no meaning.

 

[mshiddensecret]

What do you suggest?

 

[Simon Bridge]

Try saying how you arrived at the number and what it is for.
re. last line post #9

 

[mshiddensecret]

Well the formula is qv x b. So by sin 66, it makes it orthogonal to it.

 

[Simon Bridge]

So "24" was intended to be an angle to use for "theta" in the formula.

That would be 24 degrees ... the reasoning is that if "inclination" means "angle to the vertical" then the angle to the velocity vector (which is horizontal) will be 90-minus-inclination or 90-66=34 degrees

See how that is clearer than just writing a number?

Have you tried that angle in the formula to see if it gives the right answer?

Note - I still don't see how you got 24deg.
... "by sin66, it makes it orthogonal to it" is meaningless. What are the "it"s in that sentence?

Did you try any of the other suggestions in post #11?

 

[Simon Bridge]

OK - now I think I've got a handle on how you talk... I think we are getting further off track.

Back up: what is the vertical component of the Earth's magnetic field?
If the angle to the horizontal is 66deg, then what is the total field?

i.e. what is the relationship between the total field and the vertical component of the field?

 

[mshiddensecret]

Vertical is the one I found. So, if I cos 66, I should find the total field.