Is My Solution for Magnetic Field and Force Correct?

AI Thread Summary
The magnetic force calculation using F = qvB is confirmed to be correct, but the need for a sine component depends on the angle between the velocity and magnetic field. The radius of curvature calculation also appears accurate. However, the time spent calculation needs revision, as it mistakenly uses the full circumference instead of the arc length. Lastly, the direction of the magnetic field and the charge of the particles may require reevaluation, as it affects the expected motion and distance calculations. Overall, the solution has some correct elements but requires adjustments based on the problem's context.
bckcookie
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Homework Statement



As attached file! Please advise me my solution is right or not! Thanks a lot!

Homework Equations



F=qvB - (1)
R=mv/eB
d/t = v
length of arc = 2pi(R)
sinθ=a/c

The Attempt at a Solution



a) the magnetic force,
F = qvb
=1.6x10^-19 x 10^7 x 0.03N
= 4.8x10^-14N
(does it need to product with sin 45degree?? please advise!)

b) the radius of curvature
R = mv/eB
= (9.11 x 10^-31kg x 10^7) / 1.6x10^-19 x 0.03
=1.90 x 10^-3m

c) The time spend
=The length of arc / v
= 2∏R / v
= 1.194 x 10^-9s

d) the distance of x
= 2Rsin45
= 2.687 x 10^-3 m
 

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bckcookie said:

Homework Statement



As attached file! Please advise me my solution is right or not! Thanks a lot!

Homework Equations



F=qvB - (1)
R=mv/eB
d/t = v
length of arc = 2pi(R)
sinθ=a/c

The Attempt at a Solution



a) the magnetic force,
F = qvb
=1.6x10^-19 x 10^7 x 0.03N
= 4.8x10^-14N
'Looks right! :approve:
(does it need to product with sin 45degree?? please advise!)
Think about what the cross product means. The magnitude of \vec A \times \vec B is the full AB only if \vec A and \vec B are perpendicular. Is the velocity perpendicular to the magnetic field? (Hint: you need to think in three dimensions. :wink:)
b) the radius of curvature
R = mv/eB
= (9.11 x 10^-31kg x 10^7) / 1.6x10^-19 x 0.03
=1.90 x 10^-3m
'Looks right. :approve:
c) The time spend
=The length of arc / v
= 2∏R / v
= 1.194 x 10^-9s
Try that one again. 2πR is the circumference of an entire circle of radius R, not just an arc length.
d) the distance of x
= 2Rsin45
= 2.687 x 10^-3 m
'Looks right to me. :approve:
 
By the way, I just noticed that something is not quite right with the problem statement. Either that, or it's a "trick" question. You might want to tell you instructor about this.

According to attached figure, the magnetic flux density is going into the page (denoted by the 'x's). That, combined with the direction and orientation of the arrows, and the fact that electrons are negatively charged, the electrons would not curl around in a counter-clockwise direction as the problem indicates. Instead, they would curl around in a clockwise direction, and exit the magnetic field at some distance x *below* where they entered -- not above.

Either the figure needs to be redrawn to show the magnetic field coming out of the page, OR the magnetic flux density should be changed from 0.03 T to -0.03 T, OR the direction of the arrows need be to be reversed, OR the electrons should be replaced with positrons.

Keeping everything as it is, x will end up being negative, thus you'll have to redo parts c) and d).
 
you are right! Thanks for your hints!
 
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