# Magnetic field & force

1. May 20, 2012

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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

#### Attached Files:

• ###### magnetic field.jpg
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2. May 21, 2012

### collinsmark

'Looks right!
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. )
'Looks right.
Try that one again. 2πR is the circumference of an entire circle of radius R, not just an arc length.
'Looks right to me.

3. May 21, 2012

### collinsmark

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).

4. May 21, 2012