Magnetic Field of a Rectangular Loop at Point P

[says]

Homework Statement

Find the magnetic field at the point distance R outside a rectangular loop of sides 2R, 4R carrying current I as shown in the attached diagram

Homework Equations

Biot-Savart Law

The Attempt at a Solution

Could I just break this up into 4 separate lines and then calculate using the Biot-Savart Law for a finite wire? The magnetic field at the point, P, would then be the sum of the magnetic fields due to each line, no?

 

[TSny]

Could I just break this up into 4 separate lines and then calculate using the Biot-Savart Law for a finite wire? The magnetic field at the point, P, would then be the sum of the magnetic fields due to each line, no?

Yes.

 

[says]

For the top line of the rectangle:

dB = (μ₀I / 4π) ⋅ dl x r / r²

∫ dB = (μ₀ I / 4π) ⋅ ∫ dl sinθ / r²

B = (μ₀ I / 4π) ∫ (dl / R² + x²) ⋅ (R / √R²+x²)

B = (μ₀ I R / 4π) ∫ 1 / (R²+x²)^3/2 dl

bounds of integration for the top line of the rectangle are from R=(0,4)

B = μ₀I/4πR (4 / √R² + 4²)

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For the bottom line of the rectangle everything is the same except R is now 3R because it is further away.

B = μ₀I/4π3R (4 / √3R² + 4²)I've essentially followed this YouTube video to calculate B for the top and bottom line:

 

[TSny]

For the top line of the rectangle:

dB = (μ₀I / 4π) ⋅ dl x r / r²

∫ dB = (μ₀ I / 4π) ⋅ ∫ dl sinθ / r²

B = (μ₀ I / 4π) ∫ (dl / R² + x²) ⋅ (R / √R²+x²)

B = (μ₀ I R / 4π) ∫ 1 / (R²+x²)^3/2 dl

bounds of integration for the top line of the rectangle are from R=(0,4)

Pretty close. A couple of things: (1) dl should be written as dx in the integral since you are letting x represent position along the wire. (2) The limits of integration are x = 0 to x = 4R.

B = μ₀I/4πR (4 / √R² + 4²)

This isn't quite right because of the mistake in the limits mentioned above.

Note: You listed a relevant equation that represents the result of doing the integration. So, if you understand that equation, you don't need to do the integration. (Or, you can use the equation to check your result for the integration.)

In the denominator of you answer, it looks like only the R² is under the square root. Use parentheses if you want to indicate that all of R² + 4 is inside the square root.

For the bottom line of the rectangle everything is the same except R is now 3R because it is further away.

B = μ₀I/4π3R (4 / √3R² + 4²)

Again, close but not quite right because of a mistake in the limits of integration.

What about the vertical sides? Does either one of those contribute to the answer?

 

[says]

Ok, so I used the equation that I mentioned in my OP.

dB_z = μ₀I / 4πz ⋅ (sinθ₂-sinθ₁)

For the top and bottom lines of the rectangle sinθ₁ = 0 so dB_z reduces to:

dB_z = μ₀I / 4πz ⋅ (sinθ₂)

For the top line of the rectangle:

z = √(R² + (4R)²)
sinθ₂ = R / z

dB_z = μ₀ I / 4πz ⋅ (R/z)

dB_z = μ₀ I R / 4πz²

 

[TSny]

You need to express the answer in terms of R, not z.

What is the meaning of Z in the formula that you listed as the relevant equation? See http://www.phys.uri.edu/gerhard/PHY204/tsl216.pdf

 

[says]

I was using z as the hypotenuse, which I think is where the confusion has come from. I've renamed the hypotenuse to h for simplicity

dB_z = μ₀ I / 4πz ⋅ (sinθ₂-sinθ₁)

z = R (for the top line of the rectangle)

For the top line of the triangle dB_z reduces to:

dB_z = μ₀ I / 4πR ⋅ (sinθ₂)

sinθ₂ = R/h

Therefore:

dB_z = (μ₀ I / 4πR) ⋅ (R/h)

dB_z = (μ₀ I / 4πh)

Expressing the answer in terms of R:

dB_z = (μ₀ I / 4π√(R² +(4R)²)

 

[TSny]

Note that θ₂ is the angle shown:

So, sinθ₂ ≠ R/h. Otherwise, I think your work is OK.

 

[says]

ohhh in my OP the YouTube I linked it uses the θ from the other corner.

sinθ₂ = 4R / h

dB_z = μ₀ I / 4πR ⋅ (4R / h)

Therefore:

dB_z = (μ₀ I 4R / 4πRh)

dB_z = (μ₀ I / πh)

Expressing the answer in terms of R:

dB_z = (μ₀ I / πh)

B_z = μ₀ I / π √(R²+(4R)²)

That's the magnetic field at the point P due to the top line of the rectangle only

 

[TSny]

sinθ₂ = 4R / h

Yes

B_z = μ₀ I / π √(R²+(4R)²)

That's the magnetic field at the point P due to the top line of the rectangle only

OK. I would simplify the square root expression.

 

[says]

simplified:
B_z = μ₀ I / π [R²+(4R)²]^3/2

 

[TSny]

simplified:
B_z = μ₀ I / π [R²+(4R)²]^3/2

I don't see that. I just meant that you can simplify √(R² + (4R)²) to R√17.

 

[says]

For the line 2R on the right hand side of the rectangle:

sinθ₂ = 4R / h₂

sinθ₁ = 4R / h₁

dB_z = μ₀ I / 4π3R ⋅ [4R / h₂ - 4R / h₁ ]

h₂ = 4R / √((4R)²+(3R)²) = 4R / √ 25R²

h₁ = 4R / √((4R)²+(R)²) = 4R / √ 17R²

∴ dB_z = μ₀ I / 4π3R ⋅ [ 4R / √ 25R² - 4R / √ 17R² ]

 

[TSny]

For the right hand side of length 2R, the meaning of Z, θ₁, and θ₂ are as shown

 

[says]

sinθ₂ = 3R / √((4R)² + (3R)²)

sinθ₁ = R / √((4R)² + (R)²)

dB_z = μ₀ I / 4π4R ⋅ [sinθ₂ - sinθ₁]

dB_z = μ₀ I / 16πR ⋅ [3R / √((4R)² + (3R)²) - R / √((4R)² + (R)²)]

 

[says]

I'm not sure what to do for the left hand side of the rectangle. If both sinθ cancel, do i integrate 'R' from 3R to 0?

 

[TSny]

sinθ₂ = 3R / √((4R)² + (3R)²)

sinθ₁ = R / √((4R)² + (R)²)

dB_z = μ₀ I / 4π4R ⋅ [sinθ₂ - sinθ₁]

dB_z = μ₀ I / 16πR ⋅ [3R / √((4R)² + (3R)²) - R / √((4R)² + (R)²)]

I think that's correct.

 

[TSny]

I'm not sure what to do for the left hand side of the rectangle. If both sinθ cancel, do i integrate 'R' from 3R to 0?

What are the values of θ₁ and θ₂ for this side? What does the formula for B give?

Or, apply the Biot-Savart law to an infinitesimal length dl of this side of the rectangle.