Magnetic Force with Parallel currents

AI Thread Summary
The discussion revolves around calculating the magnetic field at a point near two parallel wires carrying currents in opposite directions. The magnetic field contributions from each wire are expressed using the formula B=μo*I / 2πR, leading to separate calculations for each wire. The participants confirm that the initial steps in the calculation are correct, but there is uncertainty regarding the final simplification, which yields different results than the textbook answer. Concerns are raised about the validity of the textbook answer, particularly when considering limits where one distance is much larger than the other. Ultimately, the consensus is that the logic in the calculations is sound, but the final answer may differ from the book's due to a potential error in the textbook.
Ignitia
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Homework Statement


A circuit with current I has two long parallel wire sections that carry current in opposite directions. Find magneticfieldatapointPnearthesewiresthatisadistance a from one wire and b from the other wire as shown in the figure.
Rfq8TXpsSAspAbOVLmbWEmTJjR6Ch4mY58sLIzdEgq27Vi71B5IowR~2HwbeoM_&Key-Pair-Id=APKAJ5Y6AV4GI7A555NA.jpg


Homework Equations



B=μo*I / 2πR

The Attempt at a Solution



Field for I1:
B1 = μo*I / 2πb
Field for I1:
B2 = μo*I / 2πa

First have to calculate the x and y components:

Bnetx = B2 [ - sin(θ)] + B1 [cos(0)] = B2 [ - sin(θ)] + B1
Bnety = B2 [cos(θ)] + B1 [sin(0)] = B2 [cos(θ)]

cos(θ) = sqrt(a2 - b2) / a
sin(θ) = b/a

And find the total net value, Bnet = sqrt( Bnetx2 + Bnety2) so,

Bnet = sqrt( ( (μo*I / 2πa) [ - b/a] + (μo*I / 2πb))2 + (( μo*I / 2πa) [sqrt(a2 - b2) / a])2 )

And I just simplify this whole mess, correct? If it's correct up to here, then my mistake is only in the simplifying, and not the steps leading up to it.
 

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Ignitia said:

Homework Statement


A circuit with current I has two long parallel wire sections that carry current in opposite directions. Find magneticfieldatapointPnearthesewiresthatisadistance a from one wire and b from the other wire as shown in the figure.
View attachment 222953

Homework Equations



B=μo*I / 2πR

The Attempt at a Solution



Field for I1:
B1 = μo*I / 2πb
Field for I1:
B2 = μo*I / 2πa

First have to calculate the x and y components:

Bnetx = B2 [ - sin(θ)] + B1 [cos(0)] = B2 [ - sin(θ)] + B1
Bnety = B2 [cos(θ)] + B1 [sin(0)] = B2 [cos(θ)]

cos(θ) = sqrt(a2 - b2) / a
sin(θ) = b/a

And find the total net value, Bnet = sqrt( Bnetx2 + Bnety2) so,
Correct up to here, didn't check the last line. Actually your answer is Bnetx and Bnety since B is a vector with x and y components. Your way calculates the magnitude of the net B field but actually that was not explicitly asked for.Bnet = sqrt( ( (μo*I / 2πa) [ - b/a] + (μo*I / 2πb))2 + (( μo*I / 2πa) [sqrt(a2 - b2) / a])2 )

And I just simplify this whole mess, correct? If it's correct up to here, then my mistake is only in the simplifying, and not the steps leading up to it.[/QUOTE]
 
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Likes Ignitia, Charles Link and harambe
Your steps are correct
 
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Likes Ignitia, Hydrous Caperilla and Charles Link
It looks correct to me. What answer did they give?
 
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Charles Link said:
It looks correct to me. What answer did they give?

The answer was

B = μo*I*a/2πb2

I got it down to

B = μo*I/2π * sqrt(a2-3b2)/ab

So if everyone says what I posted is correct, then the logic is fine and I just have to find the mistake leading up to the answer.
 
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Ignitia said:
The answer was

B = μo*I*a/2πb2

I got it down to

B = μo*I/2π * sqrt(a2-3b2)/ab

So if everyone says what I posted is correct, then the logic is fine and I just have to find the mistake leading up to the answer.
Edit: I get ## B=(\frac{\mu_oI }{2 \pi})(\frac{\sqrt{a^2-b^2}}{ ab}) ##. Unless we all made the same mistake, I think the book just might have an incorrect answer. ## \\ ## Also the book's answer makes no sense. When ## a ## is large, their answer becomes large. ## \\ ## For large ## a ##, you get ## B=\frac{\mu_o I}{2 \pi \, b } ## , just as you should. (The ## 3b^2 ## doesn't work for ## a \approx b ##. I originally had the same answer too, until I double-checked the last step).
 
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Charles Link said:
Edit: I get ## B=(\frac{\mu_oI }{2 \pi})(\frac{\sqrt{a^2-b^2}}{ ab}) ##. Unless we all made the same mistake, I think the book just might have an incorrect answer. ## \\ ## Also the book's answer makes no sense. When ## a ## is large, their answer becomes large. ## \\ ## For large ## a ##, you get ## B=\frac{\mu_o I}{2 \pi \, b } ## , just as you should. (The ## 3b^2 ## doesn't work for ## a \approx b ##. I originally had the same answer too, until I double-checked the last step).

The book could be wrong - it's happened before. Thanks for checking.
 
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