Solving for a neutralizing B0 in a Thomson Cathode Ray (Algebra?)

In summary, the conversation is about a problem with units not checking out in a algebraic solution. The equations and work done so far have been attached for reference. The issue is resolved when it is realized that the units are actually equivalent and the mistake was in not raising everything to the 1/2 power. The conversation ends with the person thanking for the help and mentioning that they have learned about the importance of units.
  • #1
garthenar
35
8
Homework Statement
In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the plates (with no magnetic field present). When you add the magnetic field as described in Part C, to what value do you have to adjust its magnitude B0 to observe no deflection?

Assume that the plates are 6.00 cm long and that the distance between them and the screen is 12.0 cm.
Express your answer numerically in tesla.
Relevant Equations
deltaY = eE0d(d+2L)/2m(v0)^2

V0 = E0/B0
I've attached a picture from the problem, the given equations, and my work so far.

I knew I had a problem when My units wouldn't check out. This should just be simple algebra but It's not working. Please help me to understand what I am doing wrong.
 

Attachments

  • Fig1 from Thomeson Experement.PNG
    Fig1 from Thomeson Experement.PNG
    6.1 KB · Views: 250
  • thomson exp ptD.pdf
    222 KB · Views: 328
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  • #2
I haven't gone through your whole attempted solution yet, but regarding your units,

[itex] \mathrm{T = \frac{kg}{s \cdot C} = \frac{N \cdot s}{m \cdot C}} [/itex]

These are all the same units, equivalent to 1 Tesla. You've indicated near the end of your work that they are not the same, but they in fact are the same. If it helps, recall that

[itex] \mathrm{N = \frac{kg \cdot m}{s^2}} [/itex]
 
  • #3
This is unfortunate. I confirmed what you told me about the units but my answer is still wrong.

I have double checked the values they gave me, adjusted the resolution of my constants (eg. the charge and mass of the electron) and even tried changing signs. My answer makes sense to me but is apparently wrong. And the units agree with the setup being right so I believe it has to be my values

... I forgot to raise everything to the 1/2 power... Well at least I learned something valuable about the units of magnetic field. I'll make sure to add that to my note cards. (Units are life)

Thank you for your help and have a good night.
 
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1. What is a neutralizing B0 in a Thomson Cathode Ray?

A neutralizing B0 in a Thomson Cathode Ray refers to the magnetic field that is used to counteract the electric field in the cathode ray tube, allowing for more precise and accurate measurements of electron beams.

2. How is the neutralizing B0 calculated in a Thomson Cathode Ray?

The neutralizing B0 is calculated using the formula B0 = (2V)/r^2, where V is the voltage applied to the deflecting plates and r is the radius of the cathode ray tube.

3. What is the purpose of solving for a neutralizing B0 in a Thomson Cathode Ray?

Solving for a neutralizing B0 is important in order to accurately measure the deflection of an electron beam in a cathode ray tube. Without a neutralizing B0, the electric field would overpower the magnetic field and make it difficult to measure the electron beam's path.

4. How does the neutralizing B0 affect the path of the electron beam in a Thomson Cathode Ray?

The neutralizing B0 acts as a counterbalance to the electric field, allowing for a more stable and controlled path of the electron beam. This makes it easier to measure and analyze the beam's characteristics.

5. Are there any limitations to using a neutralizing B0 in a Thomson Cathode Ray?

While a neutralizing B0 is useful in improving the accuracy of measurements in a Thomson Cathode Ray, it is not a perfect solution. Factors such as external magnetic fields and the shape of the cathode ray tube can still affect the path of the electron beam. Additionally, the formula for calculating B0 assumes a uniform electric field, which may not always be the case.

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