Four very long, current carrying wires in the same plane

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SUMMARY

The discussion focuses on calculating the current required to make the magnetic field at the center of a square formed by four long, current-carrying wires equal to zero. The currents are 8.0 A, 20.0 A, and 10.0 A, with the unknown current denoted as I. Using the formula B = μ0I/2πr, where μ0 = 4π*10^-7 and r = 0.155 m, the solution reveals that I must equal -2 A, leading to a magnitude of 2 A. The right-hand rule is clarified, indicating that the direction of the magnetic field varies based on the position of the observer relative to the wires.

PREREQUISITES
  • Understanding of magnetic fields generated by current-carrying wires
  • Familiarity with the right-hand rule for determining magnetic field direction
  • Knowledge of the Biot-Savart Law and its application
  • Basic algebra for solving equations involving magnetic fields
NEXT STEPS
  • Study the Biot-Savart Law for detailed understanding of magnetic fields
  • Practice using the right-hand rule with various current configurations
  • Explore the implications of superposition in magnetic fields
  • Learn about the effects of wire orientation on magnetic field strength
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penguinnnnnx5
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Homework Statement


Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 31.0 cm, as shown in the figure (Figure 1) . The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I.

Find the magnitude of the current that will make the magnetic field at the center of the square equal to zero.

Picture here:
http://session.masteringphysics.com/problemAsset/1003626/14/yf_Figure_28_35.jpg

Homework Equations



B = μ0I/2πr

The Attempt at a Solution



I know the steps and solution, but what I'm not clear about is how the right hand rule is used to determine the direction of the magnetic field.

Where exactly do you position your hand? On top of the wire? To the left of it? To the right of it? How much do you curl? All the way around?

Here is the solution:

μ0I/2πr where r = distance of B from the wire.

r = .5 ( 31cm) = .155m
μ0=4π*10^-7

ƩB = 0 = μ0/2πr * (I - 10 - 8 + 20)
0 = I - 10 - 8 + 20 = I + 2
I = -2A

Magnitude of I = abs(I) = 2A

So using the right hand rule, why is 20 positive whereas 10 and 8 are negative? Assuming that into the page/monitor is negative, shouldn't the B on the 20A also go into the page/monitor via RHR?
 
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hi penguinnnnnx5! :smile:
penguinnnnnx5 said:
… what I'm not clear about is how the right hand rule is used to determine the direction of the magnetic field.

your thumb points along the current

your fingers curl round the wire in the same direction that the circular magnetic field lines curl round the wire :wink:
So using the right hand rule, why is 20 positive whereas 10 and 8 are negative?

the point is to the left of the 20 but to the right of the 10 and 8

so your fingers are coming out of the page through that point for the 20, but going into the page for the 10 and 8 :smile:
 
tiny-tim said:
hi penguinnnnnx5! :smile:your thumb points along the current

your fingers curl round the wire in the same direction that the circular magnetic field lines curl round the wire :wink:the point is to the left of the 20 but to the right of the 10 and 8

so your fingers are coming out of the page through that point for the 20, but going into the page for the 10 and 8 :smile:

buddy u didnt answer dang
 
freeform2745 said:
buddy u didnt answer dang

This thread is almost 10 years old.

It was the responsibility of the original poster to ask further questions if tiny-tim's answer was not sufficiently clear to the poster. I can assure you that tiny-tim would have artfully guided the poster to a complete understanding of the solution. tiny-tim was one of the best homework helpers that Physics Forums has ever had.
 
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