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

Respuesta :

Answer:

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

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

[tex]\omega=2\pi f[/tex]

Put the value into the formula

[tex]\omega=2\times\pi\times60[/tex]

[tex]\omega=376.9\ rad/s[/tex]

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

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

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

[tex]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}}[/tex]

[tex]B_{max}=2.901\times10^{-13}\ T[/tex]

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