What is the magnitude of the field?

In summary, a student is struggling with their IB Physics homework and needs help with a problem involving a TV picture tube and a transverse magnetic field. They used the equations V= U/q and r= (mv)/(Bq) to solve for the magnetic field, but are unsure if they found the velocity correctly. Another person suggests using the equation KE = 1/2 mv^2 to find the velocity and also reminds the student to check the signs of the charge and potential difference.
  • #1
shikagami
42
0
I am really trying hard to pass my IB Physics class, but the homework is just too hard. Please help me with this problem.

In a TV picture tube, an electron in the beam is accelerated by a potential difference of 20,000 V. Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.12 m. What is the magnitude of the field?

I am not quite sure how to do this, but this is my approach... First... I used V= U/q to find the velocity of the electrons. Then I used r= (mv)/(Bq) to solve for the magnetic field. And I got 3.98 E-3 Tesla.

Thank you in advance.
 
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  • #2
Do you know the answer? I know the magnetic field part is right, but I am not sure that you found the velocity correctly.
 
  • #3
whozum said:
Do you know the answer? I know the magnetic field part is right, but I am not sure that you found the velocity correctly.


Sorry, I don't know the answer. I wasn't sure about the velocity part either. Because I used U=1/2 mv^2, which is for springs, i think. How did u solve it? And the V= U/q is for volts.
 
  • #4
Correct me if I'm wrong, I'm rusty, but isn't voltage the energy per unit charge? So if you multiply by the charge you should have the total energy, and from there using

[tex] KE = \frac{1}{2} mv^2 [/tex] solved for v you can find the velocity.
 
  • #5
So what you're trying to say is that since I know the voltage, I should multiply it by the charge of an electron (1.60 x 10^-19 C). By doing so, I will get the total energy, which I can use in the kinetic energy equation to solve for the velocity. That sounds great. So there's no need for me to use the volts equation V= U/q. Thank you very much. You have been very helpful.
 
  • #6
V=delta U/q=change in energy/charge...you were using it all along.



change in energy=-delta KE=delta U
 
  • #7
I meant delta V.

also be sure to check the signs
 

Related to What is the magnitude of the field?

1. What is the meaning of "magnitude of the field" in science?

The magnitude of the field refers to the strength or intensity of a particular physical field, such as an electric field or magnetic field. It is a measure of the force exerted by the field on a unit of charge or mass.

2. How is the magnitude of a field calculated?

The magnitude of a field can be calculated by dividing the force exerted by the field on a test charge or object by the magnitude of the test charge or object. In some cases, it can also be calculated by using mathematical equations that describe the behavior of the field.

3. What are the units of measurement for the magnitude of a field?

The units of measurement for the magnitude of a field depend on the type of field being measured. For example, the units of electric field magnitude are newtons per coulomb (N/C), while the units of magnetic field magnitude are teslas (T).

4. How does the magnitude of a field affect its behavior?

The magnitude of a field directly affects its behavior, as it determines the strength of the force exerted on charged or magnetized objects within the field. A higher magnitude of field will result in a stronger force, while a lower magnitude will result in a weaker force.

5. Can the magnitude of a field change over time?

Yes, the magnitude of a field can change over time due to various factors, such as the movement of charged particles or changes in the source of the field. In some cases, the magnitude of a field may also be intentionally manipulated by scientists or engineers for specific purposes.

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