Is My Biot-Savart Law Solution Method Correct?

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Homework Help Overview

The discussion revolves around the application of the Biot-Savart Law to determine the magnetic field generated by a current-carrying loop. The original poster attempts to analyze the contributions to the magnetic field from different sections of the loop, specifically focusing on the curved sections while dismissing the flat edges.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the correctness of the original poster's method, questioning the treatment of angles in degrees versus radians and the representation of the arc length in the integral. There is also a discussion about the contributions of different sections of the loop to the magnetic field.

Discussion Status

Some participants express agreement with the original poster's method and direction of the magnetic field, while others raise questions about the calculations and assumptions made, particularly regarding the use of degrees in the integrals. The conversation reflects a mix of validation and inquiry into the details of the approach.

Contextual Notes

There is a noted uncertainty about the correct representation of angles in the calculations, as well as the implications of using degrees instead of radians in the context of the Biot-Savart Law.

toothpaste666
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Homework Statement



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Homework Equations


dB = (μI/4π)(dLsinθ/r^2)

The Attempt at a Solution



the flat edges of the loop will not contribute to the magnetic field because sinθ = . Only the curved outer loop with radius I will call r2 and length L2 and inner loop with radius r1 and length L1 will contribute.

B = \int_{0}^{L_2} \frac{μI(dL)}{4π(r_2)^2} + \int_{L_1}^{0} \frac{μI(dL)}{4π(r_1)^2}

B = \int_{0}^{270°} \frac{μI(r_2dθ)}{4π(r_2)^2} + \int_{270°}^{0} \frac{μI(r_1dθ)}{4π(r_1)^2}

B = \frac{μI}{4π}(\int_{0}^{270°} \frac{dθ}{r_2} + \int_{270°}^{0} \frac{dθ}{r_1})

B = \frac{μI}{4π}(\frac{270°}{r_2} + (-\frac{270°}{r_1}))

B = \frac{μI(270°)}{4π}(\frac{1}{r_2} - \frac{1}{r_1})

B = \frac{μI(270°)}{4π}(\frac{1}{4} - \frac{1}{2})

B = \frac{μ(.2)(270°)}{4π}( - \frac{1}{4})

B = -\frac{μ(.2)(270°)}{16π}

and by the right hand rule I think the direction would be into the page. Is my method correct?
 

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toothpaste666 said:
I think the direction would be into the page. Is my method correct?

Yes, the method and the direction seem correct. I have not inspected the calculations.

( 270° = 3/2⋅π )
 
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Likes   Reactions: toothpaste666
Thank you! so not converting to radians will give me an incorrect answer?
 
You wrote three quarters of the circumference of a circle as ##\int r dL##, right ? Would you settle for 270 r or do you think it should be ##3/2\; \pi r## ?
 
I am not sure. I am guessing it should be in radians?
 
Why guess ?
 

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