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Rzqust [24]
3 years ago
14

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

se also that a sinusoidal potential difference with a maximum value and a frequency of =60 Hz is applied across the plates. Find , the maximum value of the induced magnetic field that occurs at r = R. Express your answers in terms of variables first and then plug in numbers afterword.
Physics
1 answer:
seropon [69]3 years ago
7 0

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

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