Magnetic Field strength Problem

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Homework Help Overview

The problem involves determining the magnetic field strength and direction at a specified point, utilizing the Biot-Savart Law. The context includes parameters such as velocity, charge, and distance, with a focus on vector calculations.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the cross product and the representation of vectors in component form. There are attempts to clarify the unit vector and its relation to the magnetic field calculation.

Discussion Status

Participants are actively engaging with the problem, providing feedback on vector representations and calculations. There is a collaborative effort to refine the understanding of the cross product and its application in this context.

Contextual Notes

Some participants express uncertainty about the correctness of their vector calculations and the necessary steps to compute the magnetic field. There are references to specific values and relationships that are being scrutinized for accuracy.

boozi
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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.
 

Attachments

  • p32-08alt1.gif
    p32-08alt1.gif
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  • Savart.png
    Savart.png
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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.
 
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?
 
boozi said:
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.
 
boozi said:
[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.)
 
Doc Al said:
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.
 
And when I'm crossing the two, I'll just need to multiply v by r and then by sin 45, right?
 
boozi said:
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.
 
Last edited:

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