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

[jforce93]

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

 

[PeterO]

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.

 

[jforce93]

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

 

[PeterO]

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

 

[rwisz]

I appreciate your approach to solving this problem and your acknowledgement of potential incorrect answers from the online practice tool. Your solution is correct and your calculation shows a good understanding of the equation and how to manipulate it to solve for the velocity of the charged particle. Well done! It is always important to double check your answers and be aware of potential sources of error. Keep up the good work!