Biot-Savart Law method problem

[toothpaste666]

Homework Statement

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.

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

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

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

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

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

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

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

[itex] B = -\frac{μ(.2)(270°)}{16π}[/itex]

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

 

[Hesch]

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 )

 

[toothpaste666]

Thank you! so not converting to radians will give me an incorrect answer?

 

[BvU]

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## ?

 

[toothpaste666]

I am not sure. I am guessing it should be in radians?

 

[BvU]

Why guess ?