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

In summary: Yes.So what is the angle between the field vector and the horizontal?So how does that relate to 66deg?I don't know.It's 66deg because it is given in the problem statement.Where is the horizontal component of the field?If the field makes an angle of 66deg with the horizontal, where is the horizontal component of the field?If the field makes an angle of 66deg with the horizontal, where is the horizontal component of the field?The horizontal component of the field is everywhere ... it is a vector field.But if you mean "where is the horizontal component of the field vector" then the answer is "everywhere along the
  • #1
mshiddensecret
36
0

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.
 
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  • #2
(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?
 
  • #3
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
 
  • #4
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?
 
  • #5
ignore my previous post
 
  • #6
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.
 
  • #7
Simon Bridge said:
... 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?
 
  • #8
6.05e-16 N was my answer for C.
 
  • #9
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.
 
  • #10
I have no idea at this point.
 
  • #11
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.
 
  • #12
24?
 
  • #13
24?
Bare numbers have no meaning.
 
  • #14
What do you suggest?
 
  • #15
Try saying how you arrived at the number and what it is for.
re. last line post #9
 
  • #16
Well the formula is qv x b. So by sin 66, it makes it orthogonal to it.
 
  • #17
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?
 
  • #18
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?
 
  • #19
Vertical is the one I found. So, if I cos 66, I should find the total field.
 

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

1. How do you calculate the magnetic field of a current-carrying wire?

To calculate the magnetic field of a current-carrying wire, you can use the equation B = (μ0 * I) / (2π * r), where B is the magnetic field, μ0 is the permeability of free space (4π * 10^-7 T*m/A), I is the current, and r is the distance from the wire.

2. What is the right-hand rule for determining the direction of a magnetic field?

The right-hand rule states that if you wrap your right hand around a current-carrying wire with your fingers pointing in the direction of the current, your thumb will point in the direction of the magnetic field.

3. How do you calculate the magnetic field of a solenoid?

The magnetic field of a solenoid can be calculated using the equation B = μ0 * n * I, where B is the magnetic field, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.

4. What is the difference between magnetic field strength and magnetic flux density?

Magnetic field strength refers to the amount of force that a magnetic field exerts on a moving charged particle, while magnetic flux density refers to the amount of magnetic field lines passing through a specific area.

5. How does the distance from a current-carrying wire affect the strength of the magnetic field?

The strength of the magnetic field is inversely proportional to the distance from a current-carrying wire. This means that as the distance increases, the strength of the magnetic field decreases. This relationship is described by the inverse square law.

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