The above figure shows the paths of three particles moving perpendicular to a un

Q: The above figure shows the paths of three particles moving perpendicular to a un

magnetic force F = qvB sin theta sin theta = sin 90 = 1 so F = qvB F = 1.6 *10^-19 * 256.2* 0.28 F = 1.14 *10^-17 N i suppose there is some error in your calculation, recheck it as there is 10^-19 factor for charge and you have answered it as 10^15, kindly check out this

ID: 2277209 • Letter: T

Question

The above figure shows the paths of three particles moving perpendicular to a uniform magnetic field.

A) Select the color of the path which corresponds to:

Green: A particle that has no charge.                         THIS ANSWER IS CORRECT
Red: A particle that has negative charge.                    THIS ANSWER IS CORRECT
Blue: A particle that has positive charge.                    THIS ANSWER IS CORRECT

B) Now assume the particle is a proton and that its initial velocity is 256.2 m/s to the right. If the magnitude of the magnetic field is 0.28 T, what is the magnitude of the force on the particle?

I know the charge of a proton is 1.6E-19 and I am pretty sure that I use F=qBv SIN? for the formula.

F=(1.6E-19)(.28)(256.2)[SIN(1)] = 6.83E15 N WRONG ANSWER

However this answer is wrong and I'm not sure why.

Any help would be greatly appreciate! Thanks!

The above figure shows the paths of three particles moving perpendicular to a uniform magnetic field. Select the color of the path which corresponds to: A particle that has no charge. Green: A particle that has no charge. Red: A particle that has negative charge. Blue: A particle that has positive charge. Now assume the particle is a proton and that its initial velocity is 256.2 m/s to the right. If the magnitude of the magnetic field is 0.28 T, what is the magnitude of the force on the particle? I know the charge of a proton is 1.6E-19 and I am pretty sure that I use F=qBv SIN? for the formula. F=(1.6E-19)(.28)(256.2)[SIN(1)] = 6.83E15 N However this answer is wrong and I'm not sure why.

Explanation / Answer

magnetic force F = qvB sin theta

sin theta = sin 90 = 1

so F = qvB

F = 1.6 *10^-19 * 256.2* 0.28

F = 1.14 *10^-17 N

i suppose there is some error in your calculation, recheck it as there is 10^-19 factor for charge and you have answered it as 10^15, kindly check out this

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