Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also that a sinusoidal potential difference with a maximum value of 160 V and a frequency of 60 Hz is applied across the plates: that is, V = (160 V) sin[2π(60 Hz)t].(a) FindBmax(R), the maximum value of the induced magnetic field that occurs at r = R. (b) PlotBmax(r) for 0 < r < 10 cm.

Answer 1

Answer:

a) $B_{max} = 1.784*10^{-12}$

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

$V_m$ = 160 V

a) $V = V_msin(2\pi ft)$

Where $f = 60$ Hz and $V_m = 160$ V

$E =V/d= \frac{V_msin(2\pi ft)}{d}$

For $r = R$

$A = \pi R^2$

Since $\Phi_E = EA$

$\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}$

From Ampere's Law:

$\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl}$ where $I_{encl}=0$

So at $r = R$,

$B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}$

For maximum B, cos(2πft) = 1. Hence,

$B_{max}=\frac{\mu_0\epsilon_0\pi fRV_m}{d}=\frac{4\pi*10^{-7}*8.85*10^{-12}*\pi*60*0.025*160}{4.7*10^{-3}}=1.784*10^{-12}$ T

b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}$

From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}$

The plot is given in the attachment.