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.

B(max) =  3.7971 ×  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 =  

and here we take as given r = R

so A = π R²    .................2

and

ФE  = E × A

ФE =  .....................3

so use use here now  Ampere's Law that is

∫ B ds =      .....................4

and

here  is = 0  and r = R

so

     .....................5

and put here value we get

B =          .....................6

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

and we get B (max)

B(max) =     ....................7

put here all value

B(max) =      

solve it we get

B(max) =  3.7971 ×  T

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.