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

a)
`B_(max) = 1.784*10^(-12)`

Step-by-step 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= (V_msin(2\pi ft))/(d)`

For
`r = R`

`A = \pi R^2`

Since
`\Phi_E = EA`

`\Phi_E=(\pi R^2V_msin(2\pi ft) )/(d)`

From Ampere's Law:

`\int B.ds = \mu_0\epsilon_0(d\Phi_E)/(dt) + \mu_0I_(encl)` where
`I_(encl)=0`

So at
`r = R`,

`B.2\pi R = \mu_0\epsilon_0(d\Phi_E)/(dt)\\B.2\pi R = \mu_0\epsilon_0(2\pi^2fR^2V_mcos(2\pi ft))/(d)\\B = (\mu_0\epsilon_0\pi fRV_mcos(2\pi ft))/(d)`

For maximum B, cos(2πft) = 1. Hence,

`B_(max)=(\mu_0\epsilon_0\pi fRV_m)/(d)=(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 = (\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 = (\mu_0\epsilon_0\pi fRV_m)/(d)`

The plot is given in the attachment.