Respuesta :
Hi there!
a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]
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.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}[/tex]
Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
[tex]B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}[/tex]
Taking out constants from the integral:
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int ds[/tex]
Since we are integrating around an entire circle, we are integrating from 0 to 2π.
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds[/tex]
Evaluate:
[tex]B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}[/tex]
Plugging in our givens to solve for the magnetic field strength of one loop:
[tex]B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T[/tex]
Multiply by the number of loops to find the total magnetic field:
[tex]B_T = N B = 0.00631 = \boxed{6.318 mT}[/tex]
b)
Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]
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.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}[/tex]
Using the diagram, if 'z' is the point's height from the center:
[tex]r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}[/tex]
Substituting this into our expression:
[tex]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}} }[/tex]
Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds[/tex]
Evaluate:
[tex]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}}}[/tex]
Multiplying by the number of loops:
[tex]B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]
Plug in the given values:
[tex]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}[/tex]
