What is the magnitude of the field?

[shikagami]

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.

 

[whozum]

Do you know the answer? I know the magnetic field part is right, but I am not sure that you found the velocity correctly.

 

[shikagami]

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.

 

[whozum]

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

$$KE = \frac{1}{2} mv^2$$ solved for v you can find the velocity.

 

[shikagami]

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.

 

[GCT]

V=delta U/q=change in energy/charge...you were using it all along.



change in energy=-delta KE=delta U

 

[GCT]

I meant delta V.

also be sure to check the signs