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
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...
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 ,
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
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
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