Will someone please answer this question and show all of your work, please. Than

Question

Will someone please answer this question and show all of your work, please. Thank you

Help Losout Go to the previous resource in the course sequence Go to the next resource in the course sequence Show Metadata Prepare a printable document Make notes and annotations about this resource ? Provide my evaluation of this resource ovide feedback messages or contribute to the course discussion about this resource set a bookmark for this resource PHY1 31S18, Homework #5: Two wires Consider two parallel conducting wires along the direction of the z axis as shown below. Wire 1 crosses the x-axis at x- 1.60 cm and carries a current of 2.90 A out of the xy-plane of the page. Wire 2 (right) crosses the x axis at x- 1.60 cm and carries a current of 8.30 A into the xy plane. AY At which value of x is the magnetic field zero? (Hint: Careful with sign) Submit Answer Tries 0/20 Preferences on w Mark NEW posts no longer new NEW Morgan Mcginnis (mcgin2mr:cmich) Reply (Sat Feb 17 02:22:54 pm 2018 (EST) Has anyone been able to figure this one out yet? It's the only I've been struggling with& just can't figure out what I'm doing wrong. Current diseussion settings: 1. Display-Al posts 2. Not new - Once marked not NEW Change? Preferences on what is marked as NEW esc G Search or type URL 2 3 4 6 8

Explanation / Answer

Using the right-hand rule, the magnetic field from the wire at x = -1.60 cm is counter-clockwise (current upward, thumb upward, fingers curve ccw). Conversely, the magnetic field from the wire at +1.60 cm is clockwise. Between the wires, the fields are in the same direction and must add directly, so there can be no zero point in that area.

The field magnitude around each wire is I*µ0/(2**r), where r is the distance from the wire.

For |x| > 1.60:

The field magnitude from wire 1 is I1*µ0/2**(x + 1.60)
The field magnitude from wire 2 is I2*µ0/2**(x - 1.60)

The zero point occurs when these magnitudes are equal (since they are in opposite directions)
I1*µ0/2**(x + 1.60) = I2*µ0/2**(x - 1.60)
I1/(x + 1.60) = I2/(x - 1.60)
I1*(x - 1.60) = I2*(x + 1.60)
(I1 - I2)*x = 1.60*(I1 + I2)
x = 1.60*(I1 + I2)/(I1 - I2)

x= 1.60*(11.2)/(-5.4)

x = -3.32 cm

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