Hi there!
a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.
![dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}](https://tex.z-dn.net/?f=dB%20%3D%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7Bid%5Cvec%7Bl%7D%20%5Ctimes%20%5Chat%7Br%7D%7D%7Br%5E2%7D)
dB = Differential Magnetic field element
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
R = radius of loop (2.15 cm = 0.0215 m)
i = Current in loop (0.460 A)
For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.
![dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}](https://tex.z-dn.net/?f=dB%20%3D%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7Bid%5Cvec%7Bl%7D%7D%7Br%5E2%7D)
Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
![B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}](https://tex.z-dn.net/?f=B%20%3D%20%5Cint%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7Bids%7D%7BR%5E2%7D)
Taking out constants from the integral:
![B =\frac{\mu_0 i}{4\pi R^2} \int ds](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%5Cmu_0%20i%7D%7B4%5Cpi%20R%5E2%7D%20%20%5Cint%20ds)
Since we are integrating around an entire circle, we are integrating from 0 to 2π.
![B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%5Cmu_0%20i%7D%7B4%5Cpi%20R%5E2%7D%20%20%5Cint%5Climits%5E%7B2%5Cpi%20R%7D_0%20%5C%2C%20ds)
Evaluate:
![B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%5Cmu_0%20i%7D%7B4%5Cpi%20R%5E2%7D%20%20%282%5Cpi%20R-%200%29%20%3D%20%5Cfrac%7B%5Cmu_0%20i%7D%7B2R%7D)
Plugging in our givens to solve for the magnetic field strength of one loop:
![B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%284%5Cpi%20%2A10%5E%7B-7%7D%29%20%280.460%29%7D%7B2%280.0215%29%7D%20%3D%201.3443%20%5Cmu%20T)
Multiply by the number of loops to find the total magnetic field:
![B_T = N B = 0.00631 = \boxed{6.318 mT}](https://tex.z-dn.net/?f=B_T%20%3D%20N%20B%20%3D%200.00631%20%3D%20%5Cboxed%7B6.318%20mT%7D)
b)
Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
![dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}](https://tex.z-dn.net/?f=dB%20%3D%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7Bid%5Cvec%7Bl%7D%20%5Ctimes%20%5Chat%7Br%7D%7D%7Br%5E2%7D)
In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.
![dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}](https://tex.z-dn.net/?f=dB%20%3D%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7Bid%5Cvec%7Bl%7D%20sin%5Ctheta%7D%7Br%5E2%7D)
Using the diagram, if 'z' is the point's height from the center:
![r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7Bz%5E2%20%2B%20R%5E2%20%7D%5C%5C%5C%5Csin%5Cphi%20%3D%20%5Cfrac%7BR%7D%7B%5Csqrt%7Bz%5E2%20%2B%20R%5E2%7D%7D)
Substituting this into our expression:
![dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }](https://tex.z-dn.net/?f=dB%20%3D%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7Bid%5Cvec%7Bl%7D%7D%7B%28%5Csqrt%7Bz%5E2%20%2B%20R%5E2%7D%29%5E2%7D%20%7D%28%5Cfrac%7BR%7D%7B%5Csqrt%7Bz%5E2%20%2B%20R%5E2%7D%7D%29%5C%5C%5C%5CdB%20%3D%20%5Cfrac%7B%5Cmu_0%7D%7B4%5Cpi%7D%20%5Cfrac%7BiRd%5Cvec%7Bl%7D%7D%7B%28z%5E2%20%2B%20R%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%20%7D)
Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
![B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_0%20iR%7D%7B4%5Cpi%20%28z%5E2%20%2B%20R%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5Cint%5Climits%5E%7B2%5Cpi%20R%7D_0%2C%20ds)
Evaluate:
![B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_0%20iR%7D%7B4%5Cpi%20%28z%5E2%20%2B%20R%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%282%5Cpi%20R%29%5C%5C%5C%5CB%20%3D%20%5Cfrac%7B%5Cmu_0%20iR%5E2%7D%7B2%20%28z%5E2%20%2B%20R%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7D)
Multiplying by the number of loops:
![B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}](https://tex.z-dn.net/?f=B_T%3D%20%5Cfrac%7B%5Cmu_0%20N%20iR%5E2%7D%7B2%20%28z%5E2%20%2B%20R%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7D)
Plug in the given values:
![B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}](https://tex.z-dn.net/?f=B_T%3D%20%5Cfrac%7B%284%5Cpi%20%2A10%5E%7B-7%7D%29%20%28470%29%20%280.460%29%280.0215%29%5E2%7D%7B2%20%28%280.095%29%5E2%20%2B%20%280.0215%29%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5C%5C%5C%5C%20%3D%20%200.00006795%20%3D%20%5Cboxed%7B67.952%20%5Cmu%20T%7D)