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

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To find the velocity of a charged particle traveling undeflected through a crossed magnetic field, the relevant equation is E = V*B, leading to V = E/B. With E at 6852 N/C and B at 3.6 T, the calculated velocity is approximately 1903.3 m/s. However, due to significant figure considerations, the answer should be presented as 1900 m/s, as the magnetic field measurement suggests using two significant figures. The discussion emphasizes the importance of accuracy in significant figures for physics calculations. Overall, the calculations are confirmed to be correct, and the formatting for the answer is noted as an issue with the online platform.
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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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