Is My Biot-Savart Law Solution Method Correct?

AI Thread Summary
The discussion centers on the application of the Biot-Savart Law to calculate the magnetic field generated by a loop of current. The contributor correctly identifies that only the curved sections of the loop contribute to the magnetic field, while flat edges do not. The calculations involve integrating the contributions from both the outer and inner loops, leading to a final expression for the magnetic field. There is a debate about whether to use degrees or radians in the calculations, with a consensus leaning towards radians for accuracy. Overall, the method and direction of the magnetic field are affirmed as correct, but clarity on the use of radians is emphasized.
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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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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