Magnetic Force with Parallel currents

[Ignitia]

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.

Homework Equations

B=μ_o*I / 2πR

The Attempt at a Solution

Field for I1:
B1 = μ_o*I / 2πb
Field for I1:
B2 = μ_o*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( ( (μ_o*I / 2πa) [ - b/a] + (μ_o*I / 2πb))² + (( μ_o*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.

 

[rude man]

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=μ_o*I / 2πR

The Attempt at a Solution

Field for I1:
B1 = μ_o*I / 2πb
Field for I1:
B2 = μ_o*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( ( (μ_o*I / 2πa) [ - b/a] + (μ_o*I / 2πb))² + (( μ_o*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]

 

[harambe]

Your steps are correct

 

[Charles Link]

It looks correct to me. What answer did they give?

 

[Ignitia]

It looks correct to me. What answer did they give?

The answer was

B = μ_o*I*a/2πb²

I got it down to

B = μ_o*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.

 

[Charles Link]

The answer was

B = μ_o*I*a/2πb²

I got it down to

B = μ_o*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).

 

[Ignitia]

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.