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 sinu

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.

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bonexptip 5 years 2021-07-18T17:38:20+00:00 1 Answers 20 views 0

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  1. Xavia
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    2021-07-18T17:39:38+00:00
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    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

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