Magnetic Field strength Problem

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SUMMARY

The discussion focuses on calculating the magnetic field strength and direction using the Biot-Savart Law for a particle with a charge of 1.60217653 x 10^-19 C moving at a velocity of 3.0 x 10^7 m/s. The user initially miscalculated the cross product involving the unit vector \hat{r} and the velocity vector v. Correct calculations reveal that the unit vector \hat{r} should be approximately -0.02/0.028i - 0.02/0.028j, and the magnitude of the cross product is determined to be v*sin(45°). This leads to the conclusion that proper vector representation is crucial for accurate magnetic field calculations.

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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

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Try writing the vector v and \hat{r} in the \hat{x}, \hat{y} 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
\hat{r}=-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 \hat{r} 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
\hat{r}=-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 \hat{r} 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:
\hat{r}=-0.01/0.028i - 0.01/0.028j
That unit vector should be (approximately): \hat{r}=-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): \hat{r}=-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 \hat{r} x \vec{v} is just vsin45, where \hat{r} is a unit vector in the r direction.
 
Last edited:

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