What Is the Velocity of a Charged Particle in a Crossed Magnetic Field?

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Homework Help Overview

The problem involves determining the velocity of a charged particle that allows it to travel undeflected through crossed electric and magnetic fields, given specific values for the electric field (E) and magnetic field (B).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the velocity using the relationship between electric and magnetic forces, expressing the equation as E = V*B and solving for V. Some participants question the precision of the answer based on significant figures and the context of the problem.

Discussion Status

Participants have engaged in discussing the calculations presented, with some affirming the approach while also noting the importance of significant figures in the final answer. There is an acknowledgment of the limitations of the online platform regarding the format of answers.

Contextual Notes

There is a mention of the online physics practice platform being known for providing answers that may not always be accurate, which raises concerns about the reliability of the results. Additionally, the discussion touches on the constraints of not being able to use scientific notation for answers.

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Homework Statement


In the figure (see attached) what is the velocity of the charged particle to the nearest tenth of a m/s that will allow it to travel undeflected through the crossed magnetic field if E = 6852 N/C and B = 3.6 T?


Homework Equations



F = E*q + q*V*B

The Attempt at a Solution



Okay, so this is an online physics practice thing (for Giancolli) I was assigned for AP work. It's known for giving wrong (but close) answers, so don't be afraid to tell me that I'm doing this right (or wrong). But the answer I got is very close to the answer it gave me.

Anyway, here is my attempt:

E*q = q*V*B
cross out q

E = V*B

V = E/B

E = 6852 N/C
B = 3.6 T

V = 1903.3 m/s


Thanks,

Jordan
 

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jforce93 said:

Homework Statement


In the figure (see attached) what is the velocity of the charged particle to the nearest tenth of a m/s that will allow it to travel undeflected through the crossed magnetic field if E = 6852 N/C and B = 3.6 T?


Homework Equations



F = E*q + q*V*B

The Attempt at a Solution



Okay, so this is an online physics practice thing (for Giancolli) I was assigned for AP work. It's known for giving wrong (but close) answers, so don't be afraid to tell me that I'm doing this right (or wrong). But the answer I got is very close to the answer it gave me.

Anyway, here is my attempt:

E*q = q*V*B
cross out q

E = V*B

V = E/B

E = 6852 N/C
B = 3.6 T

V = 1903.3 m/s


Thanks,

Jordan

Calculations look OK. Given the original data - especially the B = 3.6 - it indicates you should only give your answer to 2 figures, so 1.9 x 103 m/s

note that 1900 m/s is not correct, because if you, even mistakenly, decide to give too many figures, those extra figures have to be correct.
 
PeterO said:
Calculations look OK. Given the original data - especially the B = 3.6 - it indicates you should only give your answer to 2 figures, so 1.9 x 103 m/s

note that 1900 m/s is not correct, because if you, even mistakenly, decide to give too many figures, those extra figures have to be correct.

Thanks!
And the online thing doesn't let us (for most questions) put in scientific notation
 
jforce93 said:
Thanks!
And the online thing doesn't let us (for most questions) put in scientific notation

In that case, 1900 might be "their" form of 1.9 x 103
 

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