Magnetic Field strength Problem

[boozi]

Homework Statement

What are the magnetic field strength and direction at the dot in Figure Ex32.8, in which v = 3.0*10^7 m/s?

Figure Ex.32.8 is attached to the this post.
r = 0.02828 m
m0/4pi = 10^-7 T
v = 3.0*10^7 m/s
q = 1.60217653*10^-19 C

Homework Equations

Biot-Savart Law (attached), can't really type it...

The Attempt at a Solution

Ok. It's probably a very simple problem and it makes me feel really bad 'cause I can't solve it... I've tried to solve it with the Biot-Savart Law (check attach) with the values I mentioned above

I'm sure I calculated the cross product wrong.. How would I calculate it in this case? Thanks in advance.

 

[Astronuc]

Try writing the vector v and [itex]\hat{r}[/itex] in the [itex]\hat{x}, \hat{y}[/itex] form, and then write the cross product.

 

[boozi]

First, thanks for replying to my post. Second, here's what I did:
vector v = 0i + 3.0*10^7j
[itex]\hat{r}[/itex]=-0.01/0.028i - 0.01/0.028j
Now, hopefully, that's correct. With that, the only thing left to do is multiply
m0/4pi * q/r, which is 2.003*10^-23 by [itex]\hat{r}[/itex] and then cross
it with vector v, right?

 

[learningphysics]

First, thanks for replying to my post. Second, here's what I did:
vector v = 0i + 3.0*10^7j
[itex]\hat{r}[/itex]=-0.01/0.028i - 0.01/0.028j
Now, hopefully, that's correct. With that, the only thing left to do is multiply
m0/4pi * q/r, which is 2.003*10^-23 by [itex]\hat{r}[/itex] and then cross
it with vector v, right?

Your r vector is just -0.02i -0.02j. Other than that everything looks good.

 

[Doc Al]

[itex]\hat{r}[/itex]=-0.01/0.028i - 0.01/0.028j

That unit vector should be (approximately): [itex]\hat{r}[/itex]=-0.02/0.028i - 0.02/0.028j
(Which is consistent with what learningphysics said about the vector r.)

 

[learningphysics]

That unit vector should be (approximately): [itex]\hat{r}[/itex]=-0.02/0.028i - 0.02/0.028j
(Which is consistent with what learningphysics said about the vector r.)

Ah yes... I apologize. you need the unit vector in the r direction, not the r vector itself.

 

[boozi]

And when I'm crossing the two, I'll just need to multiply v by r and then by sin 45, right?

 

[learningphysics]

And when I'm crossing the two, I'll just need to multiply v by r and then by sin 45, right?

You're crossing v with the unit vector in the r direction... hence it's just v*1*sin45, that gives the magnitude of the cross product.

so the magnitude of [tex]\hat{r}[/tex] x [tex]\vec{v}[/tex] is just vsin45, where [tex]\hat{r}[/tex] is a unit vector in the r direction.