# Homework Help: Biot-Savart Law problem

1. Apr 22, 2015

### toothpaste666

1. The problem statement, all variables and given/known data

2. Relevant equations
dB = (μI/4π)(dLsinθ/r^2)

3. 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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2. Apr 22, 2015

### Hesch

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

( 270° = 3/2⋅π )

3. Apr 23, 2015

### toothpaste666

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

4. Apr 23, 2015

### 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 $3/2\; \pi r$ ?

5. Apr 23, 2015

### toothpaste666

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

6. Apr 23, 2015

### BvU

Why guess ?

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