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OpenStudy (chupacabraj):
OpenStudy (chupacabraj):
I thought I could use Ampere Law? No?
I divided 1.38 x 10^-5 Tm by 4 pi x 10^-7 T/m/A = 11 A
OpenStudy (chupacabraj):
Do I have to take into account the current outside ?
Parth (parthkohli):
I3 - 12 A = 11 A then?
OpenStudy (chupacabraj):
So, the other currents don't matter?
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Parth (parthkohli):
They don't
OpenStudy (chupacabraj):
Yeah, that's the confusing part since I don't know the direction of I3? That's how you determine if you add or subtract right?
Parth (parthkohli):
Sorry, I forgot the problem. See, the circulation of the magnetic field is clockwise, so the sum of currents must be inwards.
Parth (parthkohli):
therefore the net current must be 11 A inwards... which leaves us with the only possibility that I1 is inwards (if it is outwards, both currents will be outwards and so we cannot have this possibility)
so I1 - 12 A = 11 A it is
OpenStudy (chupacabraj):
Ohh, so I3 = 23 A inwards. since it's a "larger" current, the net current is inward?
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