Answer:
a) ![B_{max} = 1.784*10^{-12}](https://tex.z-dn.net/?f=B_%7Bmax%7D%20%3D%201.784%2A10%5E%7B-12%7D)
Explanation:
Given paraeters are:
R = 25 cm
d = 4.7 mm
f = 60 Hz
= 160 V
a) ![V = V_msin(2\pi ft)](https://tex.z-dn.net/?f=V%20%3D%20V_msin%282%5Cpi%20ft%29)
Where
Hz and
V
![E =V/d= \frac{V_msin(2\pi ft)}{d}](https://tex.z-dn.net/?f=E%20%3DV%2Fd%3D%20%5Cfrac%7BV_msin%282%5Cpi%20ft%29%7D%7Bd%7D)
For ![r = R](https://tex.z-dn.net/?f=r%20%3D%20R)
![A = \pi R^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20R%5E2)
Since ![\Phi_E = EA](https://tex.z-dn.net/?f=%5CPhi_E%20%3D%20EA)
![\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}](https://tex.z-dn.net/?f=%5CPhi_E%3D%5Cfrac%7B%5Cpi%20R%5E2V_msin%282%5Cpi%20ft%29%20%7D%7Bd%7D)
From Ampere's Law:
where ![I_{encl}=0](https://tex.z-dn.net/?f=I_%7Bencl%7D%3D0)
So at
,
![B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}](https://tex.z-dn.net/?f=B.2%5Cpi%20R%20%3D%20%5Cmu_0%5Cepsilon_0%5Cfrac%7Bd%5CPhi_E%7D%7Bdt%7D%5C%5CB.2%5Cpi%20R%20%3D%20%5Cmu_0%5Cepsilon_0%5Cfrac%7B2%5Cpi%5E2fR%5E2V_mcos%282%5Cpi%20ft%29%7D%7Bd%7D%5C%5CB%20%3D%20%5Cfrac%7B%5Cmu_0%5Cepsilon_0%5Cpi%20fRV_mcos%282%5Cpi%20ft%29%7D%7Bd%7D)
For maximum B, cos(2πft) = 1. Hence,
T
b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:
![B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_0%5Cepsilon_0%5Cpi%20fR%5E2V_m%7D%7Brd%7D)
From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:
![B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_0%5Cepsilon_0%5Cpi%20fRV_m%7D%7Bd%7D)
The plot is given in the attachment.
rays ultraviolet light visible light infrared light radio waves
Answer: repeatable
Explanation: Accurate repeatable is needed in a valid experiment. I think