Find the magnetic field at the origin

[gkamal]

Homework Statement

[/B]

Homework Equations

B = u0*I / 2pi * r
that's the only equation relevant in this problem

U0 is a constant wth a value of : 4pi x10^-7

The Attempt at a Solution

okay so what i did was use the equation above and find the magnetic field (B) for I1 I3 and the x and y component of I2.Also, i used the right hand rule to determine the direction of the magnetic field for each one and I get :

I1 = 3.57x10^-5 T in the -y direction
I2 x-comp = 2,17x10^-5 T in the -y direction
I2 y-comp = 2.17x10^-5 T in the +x direction
I3 = 3.57x10^-5 in the -x direction

I tried subtracting -x direction value from the +x to find the x-component at the origin and i tried adding the 2 -y values to find the y-comonent of the magnetic field at the origin I get :

x-component : -1.4x10^-5 T
y-component : 5.74x10^-5 T

but that is said to be wrong , so please any help is appreciated

 

[TSny]

Check the values for the x and y components of the field due to I₂.

 

[gkamal]

i still get 2.17 x10 ^ -5 for both. the "r" in the x and y directions is the same which is 0.0387. and the current if I2 is 4.20A. so that's :

B= [4pi x 10^-7 * 4.20A]/[2pi*0.00387] = 2.17 x 10^-5 T for both of them

I don't see where i made a mistake

 

[TSny]

Did you use the correct value of r for I₂? How did you calculate the x and y components of B₂ from the magnitude of B₂?

 

[TSny]

For I₂ it appears that you might be trying to get B_x by doing something like B_x = μ₀I₂/ (2πy), where y is the y-component of r. But this is not correct.

 

[gkamal]

should i then use sqrt[x^2+y^2] to find the distance between I2 and the origin and then use the formula to find the magnetic field of that and then using the tangent of the angle tan[x] = 3.87cm/3.87cm and use that to determine the x and y components ?

In fact i did that and i get 4.19x10^-7 for Bx and 1.53x10^-5 , but Bx seem to be way too small so i am not sure this is right

 

[TSny]

should i then use sqrt[x^2+y^2] to find the distance between I2 and the origin and then use the formula to find the magnetic field of that and then using the tangent of the angle tan[x] = 3.87cm/3.87cm and use that to determine the x and y components ?

Yes, although I would need to see your work to make sure that you are doing it correctly.

What did you get for the magnitude of B₂?

 

[gkamal]

i got 1.54x10^-5 , but if this is right everything else is correct ? i just follow the same steps using these new values or...

 

[TSny]

i got 1.54x10^-5

OK. Then what did you get for B_2x and B_2y?
Your results for B₁ and B₃ look right.

 

[gkamal]

what i got for B2x and B2y is what i said above , the stuff using the tangent and all that and the 1.54x10^-5 is the magnitude of B2

4.19x10^-7 for Bx and 1.53x10^-5 for By

 

[TSny]

what i got for B2x and B2y is what i said above , the stuff using the tangent and all that and the 1.54x10^-5 is the magnitude of B2

4.19x10^-7 for Bx and 1.53x10^-5 for By

These values aren't correct. Show how you got them.

 

[gkamal]

should i then use sqrt[x^2+y^2] to find the distance between I2 and the origin and then use the formula to find the magnetic field of that and then using the tangent of the angle tan[x] = 3.87cm/3.87cm and use that to determine the x and y components ?

In fact i did that and i get 4.19x10^-7 for Bx and 1.53x10^-5 , but Bx seem to be way too small so i am not sure this is right

i did what is mentioned above and then i used these fomulas Bx = Bsin(x) and By = Bcos(x)

 

[TSny]

You have the right method but you must be making a numerical error somewhere. It's a little confusing to use the symbol x for x-component and also for an angle.

I can't identify the error without seeing your explicit calculation. What did you get for the angle x?

 

[gkamal]

i'm sorry but my calculations are a real mess and if i take a picture and post it here it won't help u in the slightest but here , i got 1.56 for the angle which is oddly small which is why i tried again and got 45 for the angle and now my values are :

Bx = 1.09 x 10^-5
By = 1.09 x 10^-5

 

[TSny]

got 45 for the angle and now my values are :

Bx = 1.09 x 10^-5
By = 1.09 x 10^-5

OK, except you need to check if you have the correct signs for these components.

 

[gkamal]

so By should be negative and Bx should be positive ?

 

[TSny]

so By should be negative and Bx should be positive ?

Yes.

 

[gneill]

i'm sorry but my calculations are a real mess and if i take a picture and post it here it won't help u in the slightest but here , i got 1.56 for the angle which is oddly small which is why i tried again and got 45 for the angle and now my values are :

Bx = 1.09 x 10^-5
By = 1.09 x 10^-5

You need to be able to present your work if you expect helpers to be able to help you. If your handwritten efforts are indecipherable then please take the time to type your work into the post. The editor provides tools for creating reasonably well formatted math equations, and LaTeX syntax is also available for presenting more advanced math.

 

[gkamal]

Yes.

so now i do the same steps which subtracting the -x value from the +x value and adding the 2 -y values together ?

if that was correct i get : -4.66x10^-5 T in the i direction and 4.66x10^-5 in the j direction

sorry for the late reply i had a cal final exam so i didn't have the time for physics

 

[gkamal]

You need to be able to present your work if you expect helpers to be able to help you. If your handwritten efforts are indecipherable then please take the time to type your work into the post. The editor provides tools for creating reasonably well formatted math equations, and LaTeX syntax is also available for presenting more advanced math.

yeah , sorry about that i'll do it next time

 

[TSny]

so now i do the same steps which subtracting the -x value from the +x value and adding the 2 -y values together ?

That should do it.

 

[gkamal]

That should do it.

ok so this is what i did and i got -4.66x10^-5 T in the i direction and 4.66x10^-5 in the j direction but the machine keeps telling me that I'm wrong ,

 

[TSny]

Check over your work. From previous posts, you have indicated that you have the correct x and y components of the three B fields. So, you might be making a careless error in combining the components. Watch the signs. If the grader is picky about significant figures, then you might try keeping an extra significant figure in the calculations.

If you want me to try to spot a specific error, then you will need to show your work in detail.

 

[gkamal]

I'm sorry that the image is flipped but i tried uploading the pic multiple times but it still not straight , and I'm down to my last try so please look carefully at my work cause otherwise I'm getting 0 on this questions and it counts for a lot.Thanks for your help.Btw i don't get why I had to find the vector for the magnetic field cause by I2 and then separate into x and y components instead of just find the x and y components immediatly since i already have the values for both the distance in x and in y

 

[TSny]

OK. Now I can see where there are mistakes.

How can the x-component of a vector be in the -y direction?
How can the y-component of a vector be in the +x direction?

The net $\hat{i}$ component is the sum of the individual x-components. Also, you need to check if you have the correct signs for the individual components.

In problems like this, you should include a sketch where you show the individual B fields drawn at the origin roughly to scale. You can then see from the sketch what the signs of the x and y components of each field will be. You will also be able to deduce from the diagram the signs of the net x and y components and also see roughly the relative magnitudes of the net x and y components.

 

[gkamal]

ok so Bx should be in the in the +x direction and By should be in the -y direction right ? i guess i mixed them while going fast

 

[TSny]

ok so Bx should be in the in the +x direction and By should be in the -y direction right ?

Yes.

i guess i mixed them while going fast

Sometimes going fast ends up taking more time than going slow.

 

[gkamal]

This gives me -2.48x10^-5 T in the i direction and 2.48x10^-5 in the j direction does this look correct ?

 

[TSny]

This gives me -2.48x10^-5 T in the i direction and 2.48x10^-5 in the j direction does this look correct ?

I believe one of these is correct. Check your work and see if you can discover which is wrong. Did you draw a sketch? If so, you can see that the x and y components of the result will not have the same magnitude.

 

[gkamal]

you're right it should be -2.48x10^-5 T in the i direction and -4.66x10^-5 in the j direction right?

 

[TSny]

Closer, but no cigar (yet).

 

[gkamal]

actually the -4.66x10^-5 is negative not positive i just edited it like 2 sec b4 ur reply i hope this was the mistake

 

[TSny]

actually the -4.66x10^-5 is negative not positive i just edited it like 2 sec b4 ur reply i hope this was the mistake

Yes, that was the mistake. So, your final answer is?

 

[gkamal]

-2.48x10^-5 T in the i direction and -4.66x10^-5 in the j direction

 

[TSny]

I think that's it. Be sure to include units and whatever else is required.

 

[gkamal]

Yes it is right but just to make sure the B of I3 becomes positive because it's value is negative and it is point in the -x , thus it would be point in the +x with a positive value since 2 negatives cancel right?

btw thank you so much for your help

 

[TSny]

Yes it is right but just to make sure the B of I3 becomes positive because it's value is negative and it is point in the -x , thus it would be point in the +x with a positive value since 2 negatives cancel right?

That's not how I think about it. The field $\vec{B}_3$ produced by $I_3$ is a vector. So, I would not refer to it as being positive or negative. It has a magnitude that is positive. (All vectors have positive magnitude, by definition.) However, when I draw the direction of $\vec{B}_3$ in a sketch using the right hand rule for magnetic fields, I can see that it points to the right. So, the x-component is positive. [Edited]

btw thank you so much for your help

You are welcome. Glad I could be of some help.