# What is the magnitude of the magnetic force in the figure [Solved]

The Power on a) and b) are the identical. To seek out the Power use the equation F= qvB. F= (1.6*10^-19)(2.5*10^7)(.9) = 3.6e^-12N. I Cannot clarify, however I do know that the power for a) is within the -z Route, and b) is within the -zy Route.

1.6E-19(2.5E7)(.90) = magnitude

Most of those magnetic power questions presume 90 diploma angles; simply use 1 instead of sin theta. or qvb = F.
use 1.6*10^-19 for the cost on the electron, that is asking for magnitude solely so no want for the unfavorable cost on the electron.
therefor; for query half a

So its essential to do. VSin45 to search out the size reverse the angle, this size will then be equal to the dotted line above V, as a result of 45degree sides are equal. which ought to offer you z

Most of those magnetic power questions presume 90 diploma angles; simply use 1 instead of sin theta. or qvb = F.
use 1.6*10^-19 for the cost on the electron, that is asking for magnitude solely so no want for the unfavorable cost on the electron.
therefor; for query half a

1.6E-19(2.5E7)(.90) = magnitude

The Power on a) and b) are the identical. To seek out the Power use the equation F= qvB. F= (1.6*10^-19)(2.5*10^7)(.9) = 3.6e^-12N. I Cannot clarify, however I do know that the power for a) is within the -z Route, and b) is within the -zy Route.

The magnitude is the electrical cost*velocity*Magnetic subject * sin (90) for each
So: (-1.6E-19)*(2.5 x 10^7)*(.9)*(1)

Hope this helps

Most of those magnetic power questions presume 90 diploma angles; simply use 1 instead of sin theta. or qvb = F.
use 1.6*10^-19 for the cost on the electron, that is asking for magnitude solely so no want for the unfavorable cost on the electron.
therefor; for query half a

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1.6E-19(2.5E7)(.90) = magnitude

This post is last updated on hrtanswers.com at Date : 1st of September – 2022