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  • A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R= to 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1.

2.(c) A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V AC supply with an (angular) frequency of 300 rad\: \: s ^{-1}.Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

      

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It is given that the radius of each circular plate,\ R = 6.0 \, \text{cm} \, the capacitance,\ C = 100 \, \text{pF} , the capacitor is connected to a supply of \ V = 230 \, \text{V (AC supply)} \,, and the angular frequency, \ \omega = 300 \, \text{rad/s} .

The formula for the RMS value of conduction current is given by -

\ I = V \cdot \omega \cdot C \

Substitute the values:

\\ I = 230 \cdot 300 \cdot 100 \cdot 10^{-12} \ \\I = 6.9 \cdot 10^{-6} \, \text{A} = 6.9 \, \mu\text{A} \

Thus, the RMS value of the conduction current is \( 6.9 \, \mu\text{A} \).

In this parallel plate capacitor, the conduction current will be equal to the displacement current.

Therefore, the conduction current is equal to the displacement current-

\ I_{\text{displacement}} = I_{\text{conduction}} = 6.9 \, \mu\text{A}. \

For the Amplitude of the Magnetic Field, it is given that the distance between the plates from the axis, \( r = 3.0 \, \text{cm} \).

The formula for the magnetic field is:

\ B = \frac{\mu_0 \cdot r}{2\pi R^2} \cdot I_0 \

The maximum value of current is:

\[ I_0 = 2 \cdot I \]

Substitute the values:\\B = \frac{4\pi \cdot 10^{-7} \cdot 0.03 \cdot 2 \cdot 6.9 \cdot 10^{-6}}{2\pi \cdot (0.06)^2} \\ \\ B = 1.63 \cdot 10^{-11} \, \text{T}. \

Hence, the value of the amplitude of the magnetic field is \( 1.63 \cdot 10^{-11} \, \text{T} \).

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