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help please! the force between two long parallel conductors is 15 kg/metre. the conductor spacing is 10 cm. If one conductor carries twice the current of the other, calculate the current in each conductor. help me analyze please.

Mathematics
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http://theory.uwinnipeg.ca/physics/mag/node10.html
lilMissMindset: Don't ask the same question in multiple groups. "mathematics" is the wrong one anyways. -> http://openstudy.com/code-of-conduct
if you can tell, what is the expression of force per unit length between 2 infinite parallel current carrying wires ?

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Other answers:

@MuH4hA sorry. im going to erase my other questions on other groups.
Okay :) (although - you _should_ ask physics-question in physics... but i am no mod, so idk - guess it's ok ;) )
@shubhamsrg the expression of force per unit length b/w two conductors: (F2/L)=(u0*i1*i2)/2pi*d
@stgreen i knew that buddy and i wanted @lilMissMindset to ans my query, ofcorse//
i see
yea, saw it now. (F2/L)=(u0*i1*i2)/2pi*d
so where are you stuck..just plug in values given..cant be that difficult
what value should i put in 2pi*D?
is D the distance?
D is distance between the wires (in meters)
how about permeability? what is its unit?
oops sorry µ0 = 4π×10−7 ≈ 1.2566370614…×10−6 H·m−1 or N·A−2
its henry per metre
or newton per ampere?
dont worry about unit of Uo ,, just plug in everything in meters and kgs and you should get an ans in ampere..
all conventional units..
aw, sorry, im being stupid. im having trouble with the unit. :(
no no you u0 is in newton per ampere square..while the force you have is in Kg..convert your force in newtons first
i see..yes ,you have to convert that.. use mg = F for that conversion..
aww. thank you so much! at last, i get this problem. :)
@MuH4hA i got an answer in the mathematics group.
MATHS section got a number of engineers too lady :P
yea, that is why i put my questions in multiple groups, hoping for answers. not knowing, it is wrong according to codes of conduct. im one guilty student.
uh-oh come on...its not such a CRIME lol
right. lol
thanks again. ;)

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