A) ![x=\pm \frac{A}{2\sqrt{2}}](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Cfrac%7BA%7D%7B2%5Csqrt%7B2%7D%7D)
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:
![U=\frac{1}{3}K](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B3%7DK)
Using (2) we can rewrite this as
![U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B3%7D%28E-U%29%3D%5Cfrac%7B1%7D%7B3%7DE-%5Cfrac%7B1%7D%7B3%7DU%5C%5CU%3D%5Cfrac%7BE%7D%7B4%7D)
And using (1), we find
![U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BE%7D%7B4%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7DkA%5E2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B8%7DkA%5E2)
Substituting
into the last equation, we find the value of x:
![\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B8%7DkA%5E2%5C%5Cx%3D%5Cpm%20%5Cfrac%7BA%7D%7B2%5Csqrt%7B2%7D%7D)
B) ![x=\pm \frac{3}{\sqrt{10}}A](https://tex.z-dn.net/?f=x%3D%5Cpm%20%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7DA)
In this case, the kinetic energy is 1/10 of the total energy:
![K=\frac{1}{10}E](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B10%7DE)
Since we have
![K=E-U](https://tex.z-dn.net/?f=K%3DE-U)
we can write
![E-U=\frac{1}{10}E\\U=\frac{9}{10}E](https://tex.z-dn.net/?f=E-U%3D%5Cfrac%7B1%7D%7B10%7DE%5C%5CU%3D%5Cfrac%7B9%7D%7B10%7DE)
And so we find:
![\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%20%3D%20%5Cfrac%7B9%7D%7B10%7D%28%5Cfrac%7B1%7D%7B2%7DkA%5E2%29%3D%5Cfrac%7B9%7D%7B20%7DkA%5E2%5C%5Cx%5E2%20%3D%20%5Cfrac%7B9%7D%7B10%7DA%5E2%5C%5Cx%3D%5Cpm%20%5Cfrac%7B3%7D%7B%5Csqrt%7B10%7D%7DA)
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.
Answer:
80 degress is needed to boil 200g of water from 30 degress