Answer:
a. ![B= 9.45 \times10^{-3} T](https://tex.z-dn.net/?f=B%3D%209.45%20%5Ctimes10%5E%7B-3%7D%20T)
b. ![B= 0.820 T](https://tex.z-dn.net/?f=B%3D%200.820%20T)
c. ![B= 0.0584 T](https://tex.z-dn.net/?f=B%3D%200.0584%20T)
Explanation:
First, look at the picture to understand the problem before to solve it.
a. d1 = 1.1 mm
Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:
To solve the equations we have to convert all units to those of the international system. (mm→m)
![B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\](https://tex.z-dn.net/?f=B%3D%5Cfrac%7Bu_%7B0%7DI_%7Bw%7D%7D%7B2%5Cpi%20d_%7B1%7D%7D%20%3D%5Cfrac%7B52%20%5Ctimes4%5Cpi%20%5Ctimes10%5E%7B-7%7D%20%7D%7B2%5Cpi%201.1%20%5Ctimes%2010%5E%7B-3%7D%7D%20%3D9.45%20%5Ctimes10%5E%7B-3%7D%20T%5C%5C)
μ0 is the constant of proportionality
μ0=4πX10^-7 N*s2/c^2
b. d2=3.6 mm
Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:
J: current density
c: outer radius
b: inner radius
The cilinder's current is negative, as it goes on opposite direction than the wire's current.
![J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}](https://tex.z-dn.net/?f=J%3D%20%5Cfrac%20%7B-I_%7Bc%7D%7D%7B%5Cpi%28c%5E%7B2%7D-b%5E%7B2%7D%20%20%29%20%7D%7D)
![J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}](https://tex.z-dn.net/?f=J%3D%5Cfrac%7B-36%7D%7B%5Cpi%285.33%5Ctimes10%5E%7B-5%7D-1.16%5Ctimes10%5E%7B-5%7D%29%20%7D%20%3D-274.80%5Ctimes10%5E%7B3%7D%20A%2Fm%5E%7B2%7D)
![B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\](https://tex.z-dn.net/?f=B%3D%5Cfrac%7Bu_%7B0%7D%28I_%7Bw%7D-JA_%7Bs%7D%29%7D%7B2%5Cpi%20d_%7B2%7D%20%7D%20%5C%5CA_%7Bs%7D%3D%5Cpi%20%28d_%7B2%7D%5E%7B2%7D-b%5E2%29%3D4.40%5Ctimes10%5E%7B-6%7D%20m%5E2%5C%5C)
![B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B6.68%5Ctimes10%5E%7B-5%7D%7D%7B8.14%5Ctimes10%5E%7B-5%7D%7D%20%3D0.820%20T)
c. d3=7.4 mm
Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:
![B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7Bu_%7B0%7D%28I_w-I_c%29%7D%7B2%5Cpi%20d_3%20%7D%20%3D%5Cfrac%7B2.011%5Ctimes10%5E-5%7D%7B3.441%5Ctimes10%5E%7B-4%7D%7D%20%3D0.0584%20T)
As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.
A.
Destructive interference is when two waves cancel each other out or when the crest of one wave passes through the trough of another wave.
We can solve this using Snell's Law which is represented by the equation:
sin θ₁ / sin θ₂ = n₂ / n₁
From the problem, we can substitute values and solve for the angle of refraction.
sin 19 / sin θ₂ = 1.65 / 1
θ₂ = 11.38°
The angle of refraction would be 11.38°.
The position vector can be
transcribed as:
A<span> = 6 i + y j
</span>
i <span>points in the x-direction and j points
in the y-direction.</span>
The magnitude of the
vector is its dot product with itself:
<span>|A|2 = A·A</span>
<span>102 = (6 i +
y j)•(6 i+ y j)
Note that i•j = 0, and i•i = j•j =
1 </span>
<span>100 = 36 + y2
</span>
<span>64 = y2</span>
<span>get the square root of 64 = 8</span>
<span>The vertical component of the vector is 8 cm.</span>