Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 26 mm and a plate separation of 4.0 mm. Suppose also that a sinusoidal potential difference with a maximum value of 220 V and a frequency of 76 Hz is applied across the plates; that is, V = (220 V) sin[2π(76 Hz)t]. Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.

Answer 1

Answer:

B(max) =  3.7971 × $10^{-12}$  T

Explanation:

given data

radius R = 26 mm

plate separation d = 4.0 mm

potential difference Vm = 220 V

frequency f = 76 Hz

V = (220 V) sin[2π(76 Hz)t]

solution

we know that E will be

E = V ÷ d     ............1

put here value

E = $\frac{220 \times sin(2\pi 76\times t)}{d}$  

and here we take as given r = R

so A = π R²    .................2

and

ФE  = E × A

ФE = $\frac{\pi R^2 \times 220 \times sin(2\pi 76 \times t)}{d}$  .....................3

so use use here now  Ampere's Law that is

∫ B ds = $\mu_o \times \epsilon_o \times \frac{d\Phi E}{dt} + \mu_o \times I_{encl}$      .....................4

and

here $I_{encl}$  is = 0  and r = R

so

$2B \times \pi \times R = \mu_o \times \epsilon_o \times \frac{d\Phi E}{dt}$      .....................5

and put here value we get

B =  $\frac{\mu_o \times \epsilon_o \times \pi \times f \times R \times V_m cos(2\pi f t)}{d}$        .....................6

put here value  for B maximum cos(2πft) = 1

and we get B (max)

B(max) = $\frac{\mu_o\times \epsilon_o\times \pi \times f\times R\times V_m}{d}$     ....................7

put here all value

B(max) = $\frac{4\pi \times10^{-7} \times 8.85\times 10^{-12}\times \pi \times 76 \times 0.026\times 220 }{4\times 10^{-3}}$      

solve it we get

B(max) =  3.7971 × $10^{-12}$  T